Hyperbola foci formula

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Video transcript

Sours: https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:conics/x9e81a4f98389efdf:hyperb-foci/v/foci-of-a-hyperbola

Hyperbola

FAQs on Hyperbola

What is Hyperbola in Conic Section?

A hyperbola is the locus of a point whose difference of the distances from two fixed points is a constant value. The two fixed points are called the foci of the hyperbola, and the equation of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\). Here a is called the semi-major axis and b is called the semi-minor axis of the hyperbola.

What is the Equation of Hyperbola?

The equation of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\). Here 'a' is the sem-major axis, and 'b' is the semi-minor axis. There are two standard forms of equations of a hyperbola.

How to Find the General Equation of a Hyperbola?

The equation of the hyperbola can be derived from the basic definition of a hyperbola: A hyperbola is the locus of a point whose difference of the distances from two fixed points is a constant value. Let the fixed point be P(x, y), the foci are F and F'. Then the condition is PF - PF' = 2a. The difference is taken from the farther focus, and then the nearer focus. This on further substitutions and simplification we have the equation of the hyperbola as \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\).

What is a Rectangular Hyperbola?

The hyperbola having the major axis and the minor axis of equal length is called a rectangular hyperbola. Hence we have 2a = 2b, or a = b. The equation of the rectangular hyperbola is x2 - y2 = a2.

What is the Standard Equation of Hyperbola?

There are two standard equations of the Hyperbola. These equations are given as,

  • \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\), for an hyperbola having the transverse axis as the x-axis and the conjugate axis is the y-axis.
  • \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\), for an hyperbola having the transverse axis as the y-axis and its conjugate axis is the x-axis.

What is the Eccentricity of Hyperbola?

The eccentricity of the hyperbola is greater than 1. (e > 1). The eccentricity is the ratio of the distance of the focus from the center of the ellipse, and the distance of the vertex from the center of the ellipse. The distance of the focus is 'c' units, and the distance of the vertex is 'a' units, and hence the eccentricity is e = c/a. Also here we have c2 = a2 + b2.

What is the Foci of a Hyperbola?

The hyperbola has two foci on either side of its center, and on its transverse axis. The hyperbola \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) has two foci (c, 0), and (-c, 0).

What is the Conjugate Axis of a Hyperbola?

The axis line passing through the center of the hyperbola and perpendicular to its transverse axis is called the conjugate axis of the hyperbola. The conjugate axis of the hyperbola having the equation \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) is the y-axis.

What are Asymptotes of Hyperbola?

The asymptotes are the lines that are parallel to the hyperbola and are assumed to meet the hyperbola at infinity. The equation of asymptotes of the hyperbola are y = bx/a, and y = -bx/a. The equation of pair of asymptotes of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0\).

What are the Vertices of a Hyperbola?

The vertices of a hyperbola are the points where the hyperbola cuts its transverse axis. The hyperbola has only two vertices, and the vertices of the hyperbola \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) is (a, 0), and (-a, 0) respectively.

How to Find Transverse Axis of a Hyperbola?

The transverse axis of a hyperbola is a line passing through the center and the two foci of the hyperbola. A hyperbola with an equation \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) had the x-axis as its transverse axis.

Sours: https://www.cuemath.com/geometry/hyperbola/
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Learning Outcomes

  • Derive an equation for a hyperbola centered at the origin
  • Write an equation for a hyperbola centered at the origin
  • Solve an applied problem involving hyperbolas

In analytic geometry a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other.

A hyperbola

Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points [latex]\left(x,y\right)[/latex] in a plane such that the difference of the distances between [latex]\left(x,y\right)[/latex] and the foci is a positive constant.

Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances.

As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle.

Key features of the hyperbola

In this section we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x– and y-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin.

Standard Form of the Equation of a Hyperbola Centered at the Origin

Let [latex]\left(-c,0\right)[/latex] and [latex]\left(c,0\right)[/latex] be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points [latex]\left(x,y\right)[/latex] such that the difference of the distances from [latex]\left(x,y\right)[/latex] to the foci is constant.

If [latex]\left(a,0\right)[/latex] is a vertex of the hyperbola, the distance from [latex]\left(-c,0\right)[/latex] to [latex]\left(a,0\right)[/latex] is [latex]a-\left(-c\right)=a+c[/latex]. The distance from [latex]\left(c,0\right)[/latex] to [latex]\left(a,0\right)[/latex] is [latex]c-a[/latex]. The difference of the distances from the foci to the vertex is

[latex]\left(a+c\right)-\left(c-a\right)=2a[/latex]

If [latex]\left(x,y\right)[/latex] is a point on the hyperbola, we can define the following variables:

[latex]\begin{align}&{d}_{2}=\text{the distance from }\left(-c,0\right)\text{ to }\left(x,y\right)\\ &{d}_{1}=\text{the distance from }\left(c,0\right)\text{ to }\left(x,y\right)\end{align}[/latex]

By definition of a hyperbola, [latex]\lvert{d}_{2}-{d}_{1}\rvert[/latex] is constant for any point [latex]\left(x,y\right)[/latex] on the hyperbola. We know that the difference of these distances is [latex]2a[/latex] for the vertex [latex]\left(a,0\right)[/latex]. It follows that [latex]\lvert{d}_{2}-{d}_{1}\rvert=2a[/latex] for any point on the hyperbola. The derivation of the equation of a hyperbola is based on applying the distance formula, but is again beyond the scope of this text. The standard form of an equation of a hyperbola centered at the origin with vertices [latex]\left(\pm a,0\right)[/latex] and co-vertices [latex]\left(0\pm b\right)[/latex] is [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex].

A General Note: Standard Forms of the Equation of a Hyperbola with Center (0,0)

The standard form of the equation of a hyperbola with center [latex]\left(0,0\right)[/latex] and transverse axis on the x-axis is

[latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex]

where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(\pm a,0\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(0,\pm b\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(\pm c,0\right)[/latex]
  • the equations of the asymptotes are [latex]y=\pm \dfrac{b}{a}x[/latex]

The standard form of the equation of a hyperbola with center [latex]\left(0,0\right)[/latex] and transverse axis on the y-axis is

[latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex]

where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(0,\pm a\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(\pm b,0\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(0,\pm c\right)[/latex]
  • the equations of the asymptotes are [latex]y=\pm \dfrac{a}{b}x[/latex]

Note that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

(a) Horizontal hyperbola with center [latex]\left(0,0\right)[/latex] (b) Vertical hyperbola with center [latex]\left(0,0\right)[/latex]

How To: Given the equation of a hyperbola in standard form, locate its vertices and foci.

  1. Determine whether the transverse axis lies on the x– or y-axis. Notice that [latex]{a}^{2}[/latex] is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.

a. If the equation has the form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex], then the transverse axis lies on the x-axis. The vertices are located at [latex]\left(\pm a,0\right)[/latex], and the foci are located at [latex]\left(\pm c,0\right)[/latex].

b. If the equation has the form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex], then the transverse axis lies on the y-axis. The vertices are located at [latex]\left(0,\pm a\right)[/latex], and the foci are located at [latex]\left(0,\pm c\right)[/latex].

  1. Solve for [latex]a[/latex] using the equation [latex]a=\sqrt{{a}^{2}}[/latex].
  2. Solve for [latex]c[/latex] using the equation [latex]c=\sqrt{{a}^{2}+{b}^{2}}[/latex].

Example: Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation [latex]\dfrac{{y}^{2}}{49}-\dfrac{{x}^{2}}{32}=1[/latex].

Show Solution

The equation has the form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex], so the transverse axis lies on the y-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set [latex]x=0[/latex], and solve for [latex]y[/latex].

[latex]\begin{align}1=\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32} \\[1mm] 1=\frac{{y}^{2}}{49}-\frac{{0}^{2}}{32} \\[1mm] 1=\frac{{y}^{2}}{49} \\[1mm] {y}^{2}=49 \\[1mm] y=\pm \sqrt{49}=\pm 7 \end{align}[/latex]

The foci are located at [latex]\left(0,\pm c\right)[/latex]. Solving for [latex]c[/latex],

[latex]c=\sqrt{{a}^{2}+{b}^{2}}=\sqrt{49+32}=\sqrt{81}=9[/latex]

Therefore, the vertices are located at [latex]\left(0,\pm 7\right)[/latex], and the foci are located at [latex]\left(0,9\right)[/latex].

Try It

Identify the vertices and foci of the hyperbola with equation [latex]\dfrac{{x}^{2}}{9}-\dfrac{{y}^{2}}{25}=1[/latex].

Show Solution

Vertices: [latex]\left(\pm 3,0\right)[/latex]; Foci: [latex]\left(\pm \sqrt{34},0\right)[/latex]

Writing Equations of Hyperbolas in Standard Form

Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.

Hyperbolas Centered at the Origin

Reviewing the standard forms given for hyperbolas centered at [latex]\left(0,0\right)[/latex], we see that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. Note that this equation can also be rewritten as [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex]. This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.

How To: Given the vertices and foci of a hyperbola centered at [latex]\left(0,\text{0}\right)[/latex], write its equation in standard form.

  1. Determine whether the transverse axis lies on the x– or y-axis.
    1. If the given coordinates of the vertices and foci have the form [latex]\left(\pm a,0\right)[/latex] and [latex]\left(\pm c,0\right)[/latex], respectively, then the transverse axis is the x-axis. Use the standard form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex].
    2. If the given coordinates of the vertices and foci have the form [latex]\left(0,\pm a\right)[/latex] and [latex]\left(0,\pm c\right)[/latex], respectively, then the transverse axis is the y-axis. Use the standard form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex].
  2. Find [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex].
  3. Substitute the values for [latex]{a}^{2}[/latex] and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.

Example: Finding the Equation of a Hyperbola Centered at (0,0) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices [latex]\left(\pm 6,0\right)[/latex] and foci [latex]\left(\pm 2\sqrt{10},0\right)?[/latex]

Show Solution

The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex].

The vertices are [latex]\left(\pm 6,0\right)[/latex], so [latex]a=6[/latex] and [latex]{a}^{2}=36[/latex].

The foci are [latex]\left(\pm 2\sqrt{10},0\right)[/latex], so [latex]c=2\sqrt{10}[/latex] and [latex]{c}^{2}=40[/latex].

Solving for [latex]{b}^{2}[/latex], we have

[latex]\begin{align}&{b}^{2}={c}^{2}-{a}^{2} \\ &{b}^{2}=40 - 36 && \text{Substitute for }{c}^{2}\text{ and }{a}^{2}. \\ &{b}^{2}=4 && \text{Subtract}. \end{align}[/latex]

Finally, we substitute [latex]{a}^{2}=36[/latex] and [latex]{b}^{2}=4[/latex] into the standard form of the equation, [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex]. The equation of the hyperbola is [latex]\dfrac{{x}^{2}}{36}-\dfrac{{y}^{2}}{4}=1[/latex].

Try It

What is the standard form equation of the hyperbola that has vertices [latex]\left(0,\pm 2\right)[/latex] and foci [latex]\left(0,\pm 2\sqrt{5}\right)?[/latex]

Show Solution

[latex]\dfrac{{y}^{2}}{4}-\dfrac{{x}^{2}}{16}=1[/latex]

Hyperbolas Not Centered at the Origin

Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated [latex]h[/latex] units horizontally and [latex]k[/latex] units vertically, the center of the hyperbola will be [latex]\left(h,k\right)[/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[/latex] replaced by [latex]\left(x-h\right)[/latex] and [latex]y[/latex] replaced by [latex]\left(y-k\right)[/latex].

A General Note: Standard Forms of the Equation of a Hyperbola with Center (h, k)

The standard form of the equation of a hyperbola with center [latex]\left(h,k\right)[/latex] and transverse axis parallel to the x-axis is
[latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]
where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(h\pm a,k\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex]

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is [latex]2a[/latex] and its width is [latex]2b[/latex]. The slopes of the diagonals are [latex]\pm \frac{b}{a}[/latex], and each diagonal passes through the center [latex]\left(h,k\right)[/latex]. Using the point-slope formula, it is simple to show that the equations of the asymptotes are [latex]y=\pm \frac{b}{a}\left(x-h\right)+k[/latex].

The standard form of the equation of a hyperbola with center [latex]\left(h,k\right)[/latex] and transverse axis parallel to the y-axis is
[latex]\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}-\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex]
where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(h,k\pm a\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex]

Using the reasoning above, the equations of the asymptotes are [latex]y=\pm \frac{a}{b}\left(x-h\right)+k[/latex].

(a) Horizontal hyperbola with center [latex]\left(h,k\right)[/latex] (b) Vertical hyperbola with center [latex]\left(h,k\right)[/latex]

Like hyperbolas centered at the origin, hyperbolas centered at a point [latex]\left(h,k\right)[/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.

How To: Given the vertices and foci of a hyperbola centered at [latex]\left(h,k\right)[/latex], write its equation in standard form.

  1. Determine whether the transverse axis is parallel to the x– or y-axis.
    1. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form [latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex].
    2. If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form [latex]\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}-\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex].
  2. Identify the center of the hyperbola, [latex]\left(h,k\right)[/latex], using the midpoint formula and the given coordinates for the vertices.
  3. Find [latex]{a}^{2}[/latex] by solving for the length of the transverse axis, [latex]2a[/latex] , which is the distance between the given vertices.
  4. Find [latex]{c}^{2}[/latex] using [latex]h[/latex] and [latex]k[/latex] found in Step 2 along with the given coordinates for the foci.
  5. Solve for [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex].
  6. Substitute the values for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.

Example: Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices at [latex]\left(0,-2\right)[/latex] and [latex]\left(6,-2\right)[/latex] and foci at [latex]\left(-2,-2\right)[/latex] and [latex]\left(8,-2\right)?[/latex]

Show Solution

The y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x-axis. Thus, the equation of the hyperbola will have the form

 

[latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]

First, we identify the center, [latex]\left(h,k\right)[/latex]. The center is halfway between the vertices [latex]\left(0,-2\right)[/latex] and [latex]\left(6,-2\right)[/latex]. Applying the midpoint formula, we have

[latex]\left(h,k\right)=\left(\dfrac{0+6}{2},\dfrac{-2+\left(-2\right)}{2}\right)=\left(3,-2\right)[/latex]

Next, we find [latex]{a}^{2}[/latex]. The length of the transverse axis, [latex]2a[/latex], is bounded by the vertices. So, we can find [latex]{a}^{2}[/latex] by finding the distance between the x-coordinates of the vertices.

[latex]\begin{gathered}2a=|0 - 6| \\ 2a=6 \\ a=3 \\ {a}^{2}=9 \end{gathered}[/latex]

Now we need to find [latex]{c}^{2}[/latex]. The coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex]. So [latex]\left(h-c,k\right)=\left(-2,-2\right)[/latex] and [latex]\left(h+c,k\right)=\left(8,-2\right)[/latex]. We can use the x-coordinate from either of these points to solve for [latex]c[/latex]. Using the point [latex]\left(8,-2\right)[/latex], and substituting [latex]h=3[/latex],

[latex]\begin{gathered}h+c=8 \\ 3+c=8 \\ c=5 \\ {c}^{2}=25 \end{gathered}[/latex]

Next, solve for [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}:[/latex]

[latex]\begin{align}{b}^{2}&={c}^{2}-{a}^{2} \\ &=25 - 9 \\ &=16 \end{align}[/latex]

Finally, substitute the values found for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation.

[latex]\dfrac{{\left(x - 3\right)}^{2}}{9}-\dfrac{{\left(y+2\right)}^{2}}{16}=1[/latex]

Try It

What is the standard form equation of the hyperbola that has vertices [latex]\left(1,-2\right)[/latex] and [latex]\left(1,\text{8}\right)[/latex] and foci [latex]\left(1,-10\right)[/latex] and [latex]\left(1,16\right)?[/latex]

Show Solution

[latex]\dfrac{{\left(y - 3\right)}^{2}}{25}-\dfrac{{\left(x - 1\right)}^{2}}{144}=1[/latex]

Solving Applied Problems Involving Hyperbolas

As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. For example a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide!

description in caption

Cooling towers at the Drax power station in North Yorkshire, United Kingdom (credit: Les Haines, Flickr)

The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In Example 6 we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides.

Example: Solving Applied Problems Involving Hyperbolas

The design layout of a cooling tower is shown below. The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart.

Project design for a natural draft cooling tower

Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

Show Solution

We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin: [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex], where the branches of the hyperbola form the sides of the cooling tower. We must find the values of [latex]{a}^{2}[/latex] and [latex]{b}^{2}[/latex] to complete the model.

First, we find [latex]{a}^{2}[/latex]. Recall that the length of the transverse axis of a hyperbola is [latex]2a[/latex]. This length is represented by the distance where the sides are closest, which is given as [latex]\text{ }65.3\text{ }[/latex] meters. So, [latex]2a=60[/latex]. Therefore, [latex]a=30[/latex] and [latex]{a}^{2}=900[/latex].

To solve for [latex]{b}^{2}[/latex], we need to substitute for [latex]x[/latex] and [latex]y[/latex] in our equation using a known point. To do this, we can use the dimensions of the tower to find some point [latex]\left(x,y\right)[/latex] that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y-axis bisects the tower, our x-value can be represented by the radius of the top, or 36 meters. The y-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore,

[latex]\begin{align}\dfrac{{x}^{2}}{{a}^{2}}&-\dfrac{{y}^{2}}{{b}^{2}}=1 && \text{Standard form of horizontal hyperbola}. \\ {b}^{2}&=\frac{{y}^{2}}{\frac{{x}^{2}}{{a}^{2}}-1} && \text{Isolate }{b}^{2} \\ &=\frac{{\left(79.6\right)}^{2}}{\frac{{\left(36\right)}^{2}}{900}-1} && \text{Substitute for }{a}^{2},x,\text{ and }y \\ &\approx 14400.3636 && \text{Round to four decimal places} \end{align}[/latex]

The sides of the tower can be modeled by the hyperbolic equation

[latex]\dfrac{{x}^{2}}{900}-\dfrac{{y}^{2}}{14400.3636 }=1,\text{ or }\dfrac{{x}^{2}}{{30}^{2}}-\dfrac{{y}^{2}}{{120.0015}^{2} }=1[/latex]

Try It

A design for a cooling tower project is shown below. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

Show Solution

The sides of the tower can be modeled by the hyperbolic equation. [latex]\dfrac{{x}^{2}}{400}-\dfrac{{y}^{2}}{3600}=1\text{ or }\dfrac{{x}^{2}}{{20}^{2}}-\dfrac{{y}^{2}}{{60}^{2}}=1[/latex].

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8.3: The Hyperbola

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Learning Objectives

  • Locate a hyperbola’s vertices and foci.
  • Write equations of hyperbolas in standard form.
  • Graph hyperbolas centered at the origin.
  • Graph hyperbolas not centered at the origin.
  • Solve applied problems involving hyperbolas.

What do paths of comets, supersonic booms, ancient Grecian pillars, and natural draft cooling towers have in common? They can all be modeled by the same type of conic. For instance, when something moves faster than the speed of sound, a shock wave in the form of a cone is created. A portion of a conic is formed when the wave intersects the ground, resulting in a sonic boom (Figure \(\PageIndex{1}\)).

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Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. The crack of a whip occurs because the tip is exceeding the speed of sound. The bullets shot from many firearms also break the sound barrier, although the bang of the gun usually supersedes the sound of the sonic boom.

Locating the Vertices and Foci of a Hyperbola

In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other (Figure \(\PageIndex{2}\)).

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Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points \((x,y)\) in a plane such that the difference of the distances between \((x,y)\) and the foci is a positive constant.

Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances.

As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle (Figure \(\PageIndex{3}\)).

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In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the \(x\)- and \(y\)-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin.

Deriving the Equation of an Ellipse Centered at the Origin

Let \((−c,0)\) and \((c,0)\) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points \((x,y)\) such that the difference of the distances from \((x,y)\) to the foci is constant. See Figure \(\PageIndex{4}\).

A horizontal hyperbola in the x y coordinate system centered at (0, 0) with Vertices at (negative a, 0) and (a, 0) and Foci at (negative c, 0) and (c, 0), with lines of length d1 and d2 connecting a point on the right branch of the hyperbola to the foci.

If \((a,0)\) is a vertex of the hyperbola, the distance from \((−c,0)\) to \((a,0)\) is \(a−(−c)=a+c\). The distance from \((c,0)\) to \((a,0)\) is \(c−a\). The sum of the distances from the foci to the vertex is

\((a+c)−(c−a)=2a\)

If \((x,y)\) is a point on the hyperbola, we can define the following variables:

\(d_2=\) the distance from \((−c,0)\) to \((x,y)\)

\(d_1=\) the distance from \((c,0)\) to \((x,y)\)

By definition of a hyperbola, \(d_2−d_1\) is constant for any point \((x,y)\) on the hyperbola. We know that the difference of these distances is \(2a\) for the vertex \((a,0)\). It follows that \(d_2−d_1=2a\) for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses.

\[\begin{align*} d_2-d_1&=2a\\ \sqrt{{(x-(-c))}^2+{(y-0)}^2}-\sqrt{{(x-c)}^2+{(y-0)}^2}&=2a\qquad \text{Distance Formula}\\ \sqrt{{(x+c)}^2+y^2}-\sqrt{{(x-c)}^2+y^2}&=2a\qquad \text{Simplify expressions.}\\ \sqrt{{(x+c)}^2+y^2}&=2a+\sqrt{{(x-c)}^2+y^2}\qquad \text{Move radical to opposite side.}\\ {(x+c)}^2+y^2&={(2a+\sqrt{{(x-c)}^2+y^2})}^2\qquad \text{Square both sides.}\\ x^2+2cx+c^2+y^2&=4a^2+4a\sqrt{{(x-c)}^2+y^2}+{(x-c)}^2+y^2\qquad \text{Expand the squares.}\\ x^2+2cx+c^2+y^2&=4a^2+4a\sqrt{{(x-c)}^2+y^2}+x^2-2cx+c^2+y^2\qquad \text{Expand remaining square.}\\ 2cx&=4a^2+4a\sqrt{{(x-c)}^2+y^2}-2cx\qquad \text{Combine like terms.}\\ 4cx-4a^2&=4a\sqrt{{(x-c)}^2+y^2}\qquad \text{Isolate the radical.}\\ cx-a^2&=a\sqrt{{(x-c)}^2+y^2}\qquad \text{Divide by 4.}\\ {(cx-a^2)}^2&=a^2{\left[\sqrt{{(x-c)}^2+y^2}\right]}^2\qquad \text{Square both sides.}\\ c^2x^2-2a^2cx+a^4&=a^2(x^2-2cx+c^2+y^2)\qquad \text{Expand the squares.}\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\qquad \text{Distribute } a^2\\ a^4+c^2x^2&=a^2x^2+a^2c^2+a^2y^2\qquad \text{Combine like terms.}\\ c^2x^2-a^2x^2-a^2y^2&=a^2c^2-a^4\qquad \text{Rearrange terms.}\\ x^2(c^2-a^2)-a^2y^2&=a^2(c^2-a^2)\qquad \text{Factor common terms.}\\ x^2b^2-a^2y^2&=a^2b^2\qquad \text{Set } b^2=c^2−a^2\\. \dfrac{x^2b^2}{a^2b^2}-\dfrac{a^2y^2}{a^2b^2}&=\dfrac{a^2b^2}{a^2b^2}\qquad \text{Divide both sides by } a^2b^2\\ \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}&=1\\ \end{align*}\]

This equation defines a hyperbola centered at the origin with vertices \((\pm a,0)\) and co-vertices \((0,\pm b)\).

STANDARD FORMS OF THE EQUATION OF A HYPERBOLA WITH CENTER \((0,0)\)

The standard form of the equation of a hyperbola with center \((0,0)\) and transverse axis on the \(x\)-axis is

\(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\)

where

  • the length of the transverse axis is \(2a\)
  • the coordinates of the vertices are \((\pm a,0)\)
  • the length of the conjugate axis is \(2b\)
  • the coordinates of the co-vertices are \((0,\pm b)\)
  • the distance between the foci is \(2c\), where \(c^2=a^2+b^2\)
  • the coordinates of the foci are \((\pm c,0)\)
  • the equations of the asymptotes are \(y=\pm \dfrac{b}{a}x\)

See Figure \(\PageIndex{5a}\).

The standard form of the equation of a hyperbola with center \((0,0)\) and transverse axis on the \(y\)-axis is

\(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\)

where

  • the length of the transverse axis is \(2a\)
  • the coordinates of the vertices are \((0,\pm a)\)
  • the length of the conjugate axis is \(2b\)
  • the coordinates of the co-vertices are \((\pm b,0)\)
  • the distance between the foci is \(2c\), where \(c^2=a^2+b^2\)
  • the coordinates of the foci are \((0,\pm c)\)
  • the equations of the asymptotes are \(y=\pm \dfrac{a}{b}x\)

See Figure \(\PageIndex{5b}\).

Note that the vertices, co-vertices, and foci are related by the equation \(c^2=a^2+b^2\). When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

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How to: Given the equation of a hyperbola in standard form, locate its vertices and foci

  1. Determine whether the transverse axis lies on the \(x\)- or \(y\)-axis. Notice that \(a^2\) is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.
    • If the equation has the form \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\), then the transverse axis lies on the \(x\)-axis. The vertices are located at \((\pm a,0)\), and the foci are located at \((\pm c,0)\).
    • If the equation has the form \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\), then the transverse axis lies on the \(y\)-axis. The vertices are located at \((0,\pm a)\), and the foci are located at \((0,\pm c)\).
  2. Solve for \(a\) using the equation \(a=\sqrt{a^2}\).
  3. Solve for \(c\) using the equation \(c=\sqrt{a^2+b^2}\).

Example \(\PageIndex{1}\): Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation \(\dfrac{y^2}{49}−\dfrac{x^2}{32}=1\).

Solution

The equation has the form \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\), so the transverse axis lies on the \(y\)-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set \(x=0\), and solve for \(y\).

\[\begin{align*} 1&=\dfrac{y^2}{49}-\dfrac{x^2}{32}\\ 1&=\dfrac{y^2}{49}-\dfrac{0^2}{32}\\ 1&=\dfrac{y^2}{49}\\ y^2&=49\\ y&=\pm \sqrt{49}\\ &=\pm 7 \end{align*}\]

The foci are located at \((0,\pm c)\). Solving for \(c\),

\[\begin{align*} c&=\sqrt{a^2+b^2}\\ &=\sqrt{49+32}\\ &=\sqrt{81}\\ &=9 \end{align*}\]

Therefore, the vertices are located at \((0,\pm 7)\), and the foci are located at \((0,9)\).

Exercise \(\PageIndex{1}\)

Identify the vertices and foci of the hyperbola with equation \(\dfrac{x^2}{9}−\dfrac{y^2}{25}=1\).

Answer

Vertices: \((\pm 3,0)\); Foci: \((\pm \sqrt{34},0)\)

Writing Equations of Hyperbolas in Standard Form

Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.

Hyperbolas Centered at the Origin

Reviewing the standard forms given for hyperbolas centered at \((0,0)\),we see that the vertices, co-vertices, and foci are related by the equation \(c^2=a^2+b^2\). Note that this equation can also be rewritten as \(b^2=c^2−a^2\). This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.

How to: Given the vertices and foci of a hyperbola centered at \((0,0)\), write its equation in standard form

  1. Determine whether the transverse axis lies on the \(x\)- or \(y\)-axis.
    • If the given coordinates of the vertices and foci have the form \((\pm a,0)\) and \((\pm c,0)\), respectively, then the transverse axis is the \(x\)-axis. Use the standard form \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\).
    • If the given coordinates of the vertices and foci have the form \((0,\pm a)\) and \((0,\pm c)\), respectively, then the transverse axis is the \(y\)-axis. Use the standard form \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\).
  2. Find \(b^2\) using the equation \(b^2=c^2−a^2\).
  3. Substitute the values for \(a^2\) and \(b^2\) into the standard form of the equation determined in Step 1.

Example \(\PageIndex{2}\): Finding the Equation of a Hyperbola Centered at \((0,0)\) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices \((\pm 6,0)\) and foci \((\pm 2\sqrt{10},0)\)?

Solution

The vertices and foci are on the \(x\)-axis. Thus, the equation for the hyperbola will have the form \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\).

The vertices are \((\pm 6,0)\), so \(a=6\) and \(a^2=36\).

The foci are \((\pm 2\sqrt{10},0)\), so \(c=2\sqrt{10}\) and \(c^2=40\).

Solving for \(b^2\), we have

\[\begin{align*} b^2&=c^2-a^2\\ b^2&=40-36\qquad \text{Substitute for } c^2 \text{ and } a^2\\ b^2&=4\qquad \text{Subtract.} \end{align*}\]

Finally, we substitute \(a^2=36\) and \(b^2=4\) into the standard form of the equation, \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\). The equation of the hyperbola is \(\dfrac{x^2}{36}−\dfrac{y^2}{4}=1\), as shown in Figure \(\PageIndex{6}\).

A horizontal hyperbola centered at (0, 0) in the x y coordinate system with Vertices at (negative 6, 0) and (6, 0).

Exercise \(\PageIndex{2}\)

What is the standard form equation of the hyperbola that has vertices \((0,\pm 2)\) and foci \((0,\pm 2\sqrt{5})\)?

Answer

\(\dfrac{y^2}{4}−\dfrac{x^2}{16}=1\)

Hyperbolas Not Centered at the Origin

Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated \(h\) units horizontally and \(k\) units vertically, the center of the hyperbola will be \((h,k)\). This translation results in the standard form of the equation we saw previously, with \(x\) replaced by \((x−h)\) and \(y\) replaced by \((y−k)\).

STANDARD FORMS OF THE EQUATION OF A HYPERBOLA WITH CENTER \((H, K)\)

The standard form of the equation of a hyperbola with center \((h,k)\) and transverse axis parallel to the \(x\)-axis is

\[\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\]

where

  • the length of the transverse axis is \(2a\)
  • the coordinates of the vertices are \((h\pm a,k)\)
  • the length of the conjugate axis is \(2b\)
  • the coordinates of the co-vertices are \((h,k\pm b)\)
  • the distance between the foci is \(2c\), where \(c^2=a^2+b^2\)
  • the coordinates of the foci are \((h\pm c,k)\)

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is \(2a\) and its width is \(2b\). The slopes of the diagonals are \(\pm \dfrac{b}{a}\),and each diagonal passes through the center \((h,k)\). Using the point-slope formula, it is simple to show that the equations of the asymptotes are \(y=\pm \dfrac{b}{a}(x−h)+k\). See Figure \(\PageIndex{7a}\).

The standard form of the equation of a hyperbola with center \((h,k)\) and transverse axis parallel to the \(y\)-axis is

\[\dfrac{{(y−k)}^2}{a^2}−\dfrac{{(x−h)}^2}{b^2}=1\]

where

  • the length of the transverse axis is \(2a\)
  • the coordinates of the vertices are \((h,k\pm a)\)
  • the length of the conjugate axis is \(2b\)
  • the coordinates of the co-vertices are \((h\pm b,k)\)
  • the distance between the foci is \(2c\), where \(c^2=a^2+b^2\)
  • the coordinates of the foci are \((h,k\pm c)\)

Using the reasoning above, the equations of the asymptotes are \(y=\pm \dfrac{a}{b}(x−h)+k\). See Figure \(\PageIndex{7b}\).

This is a horizontal parabola opening to the right with Vertex (0, 0), Focus (6, 0), and Directrix x = negative 6. The Latus Rectum is shown, a vertical line passing through the Focus and terminating on the parabola at (6, 12) and (6, negative 12).

Like hyperbolas centered at the origin, hyperbolas centered at a point \((h,k)\) have vertices, co-vertices, and foci that are related by the equation \(c^2=a^2+b^2\). We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.

How to: Given the vertices and foci of a hyperbola centered at \((h,k)\),write its equation in standard form

  1. Determine whether the transverse axis is parallel to the \(x\)- or \(y\)-axis.
    • If the \(y\)-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the \(x\)-axis. Use the standard form \(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\).
    • If the \(x\)-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the \(y\)-axis. Use the standard form \(\dfrac{{(y−k)}^2}{a^2}−\dfrac{{(x−h)}^2}{b^2}=1\).
  2. Identify the center of the hyperbola, \((h,k)\),using the midpoint formula and the given coordinates for the vertices.
  3. Find \(a^2\) by solving for the length of the transverse axis, \(2a\), which is the distance between the given vertices.
  4. Find \(c^2\) using \(h\) and \(k\) found in Step 2 along with the given coordinates for the foci.
  5. Solve for \(b^2\) using the equation \(b^2=c^2−a^2\).
  6. Substitute the values for \(h\), \(k\), \(a^2\), and \(b^2\) into the standard form of the equation determined in Step 1.

Example \(\PageIndex{3}\): Finding the Equation of a Hyperbola Centered at \((h, k)\) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices at \((0,−2)\) and \((6,−2)\) and foci at \((−2,−2)\) and \((8,−2)\)?

Solution

The \(y\)-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the \(x\)-axis. Thus, the equation of the hyperbola will have the form

\(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\)

First, we identify the center, \((h,k)\). The center is halfway between the vertices \((0,−2)\) and \((6,−2)\). Applying the midpoint formula, we have

\((h,k)=(\dfrac{0+6}{2},\dfrac{−2+(−2)}{2})=(3,−2)\)

Next, we find \(a^2\). The length of the transverse axis, \(2a\),is bounded by the vertices. So, we can find \(a^2\) by finding the distance between the \(x\)-coordinates of the vertices.

\[\begin{align*} 2a&=| 0-6 |\\ 2a&=6\\ a&=3\\ a^2&=9 \end{align*}\]

Now we need to find \(c^2\). The coordinates of the foci are \((h\pm c,k)\). So \((h−c,k)=(−2,−2)\) and \((h+c,k)=(8,−2)\). We can use the \(x\)-coordinate from either of these points to solve for \(c\). Using the point \((8,−2)\), and substituting \(h=3\),

\[\begin{align*} h+c&=8\\ 3+c&=8\\ c&=5\\ c^2&=25 \end{align*}\]

Next, solve for \(b^2\) using the equation \(b^2=c^2−a^2\):

\[\begin{align*} b^2&=c^2-a^2\\ &=25-9\\ &=16 \end{align*}\]

Finally, substitute the values found for \(h\), \(k\), \(a^2\),and \(b^2\) into the standard form of the equation.

\(\dfrac{{(x−3)}^2}{9}−\dfrac{{(y+2)}^2}{16}=1\)

Exercise \(\PageIndex{3}\)

What is the standard form equation of the hyperbola that has vertices \((1,−2)\) and \((1,8)\) and foci \((1,−10)\) and \((1,16)\)?

Answer

\(\dfrac{{(y−3)}^2}{25}+\dfrac{{(x−1)}^2}{144}=1\)

Graphing Hyperbolas Centered at the Origin

When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\) for horizontal hyperbolas and the standard form \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\) for vertical hyperbolas.

How to: Given a standard form equation for a hyperbola centered at \((0,0)\), sketch the graph

  1. Determine which of the standard forms applies to the given equation.
  2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes.
    • If the equation is in the form \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\), then
      • the transverse axis is on the \(x\)-axis
      • the coordinates of the vertices are \((\pm a,0)\0
      • the coordinates of the co-vertices are \((0,\pm b)\)
      • the coordinates of the foci are \((\pm c,0)\)
      • the equations of the asymptotes are \(y=\pm \dfrac{b}{a}x\)
    • If the equation is in the form \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\), then
      • the transverse axis is on the \(y\)-axis
      • the coordinates of the vertices are \((0,\pm a)\)
      • the coordinates of the co-vertices are \((\pm b,0)\)
      • the coordinates of the foci are \((0,\pm c)\)
      • the equations of the asymptotes are \(y=\pm \dfrac{a}{b}x\)
  3. Solve for the coordinates of the foci using the equation \(c=\pm \sqrt{a^2+b^2}\).
  4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.

Example \(\PageIndex{4}\): Graphing a Hyperbola Centered at \((0,0)\) Given an Equation in Standard Form

Graph the hyperbola given by the equation \(\dfrac{y^2}{64}−\dfrac{x^2}{36}=1\). Identify and label the vertices, co-vertices, foci, and asymptotes.

Solution

The standard form that applies to the given equation is \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\). Thus, the transverse axis is on the \(y\)-axis

The coordinates of the vertices are \((0,\pm a)=(0,\pm \sqrt{64})=(0,\pm 8)\)

The coordinates of the co-vertices are \((\pm b,0)=(\pm \sqrt{36}, 0)=(\pm 6,0)\)

The coordinates of the foci are \((0,\pm c)\), where \(c=\pm \sqrt{a^2+b^2}\). Solving for \(c\), we have

\(c=\pm \sqrt{a^2+b^2}=\pm \sqrt{64+36}=\pm \sqrt{100}=\pm 10\)

Therefore, the coordinates of the foci are \((0,\pm 10)\)

The equations of the asymptotes are \(y=\pm \dfrac{a}{b}x=\pm \dfrac{8}{6}x=\pm \dfrac{4}{3}x\)

Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in Figure \(\PageIndex{8}\).

A vertical hyperbola centered at (0, 0) in the x y coordinate system with Vertices at (0, 8) and (0, negative 8) and Foci at (0, negative 10) and (0, 10). Also shown are the slant asymptotes, y = (4/3)x and y = (negative 4/3)x. The points (negative 6, 0) (6, 0) and (0, 0) are labeled.

Exercise \(\PageIndex{4}\)

Graph the hyperbola given by the equation \(\dfrac{x^2}{144}−\dfrac{y^2}{81}=1\). Identify and label the vertices, co-vertices, foci, and asymptotes.

Answer

vertices: \((\pm 12,0)\); co-vertices: \((0,\pm 9)\); foci: \((\pm 15,0)\); asymptotes: \(y=\pm \dfrac{3}{4}x\);

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Graphing Hyperbolas Not Centered at the Origin

Graphing hyperbolas centered at a point \((h,k)\) other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms \(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\) for horizontal hyperbolas, and \(\dfrac{{(y−k)}^2}{a^2}−\dfrac{{(x−h)}^2}{b^2}=1\) for vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes.

How to: Given a general form for a hyperbola centered at \((h, k)\), sketch the graph

  1. Convert the general form to that standard form. Determine which of the standard forms applies to the given equation.
  2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes.
    1. If the equation is in the form \(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\), then
      • the transverse axis is parallel to the \(x\)-axis
      • the center is \((h,k)\)
      • the coordinates of the vertices are \((h\pm a,k)\)
      • the coordinates of the co-vertices are \((h,k\pm b)\)
      • the coordinates of the foci are \((h\pm c,k)\)
      • the equations of the asymptotes are \(y=\pm \dfrac{b}{a}(x−h)+k\)
    2. If the equation is in the form \(\dfrac{{(y−k)}^2}{a^2}−\dfrac{{(x−h)}^2}{b^2}=1\), then
      • the transverse axis is parallel to the \(y\)-axis
      • the center is \((h,k)\)
      • the coordinates of the vertices are \((h,k\pm a)\)
      • the coordinates of the co-vertices are \((h\pm b,k)\)
      • the coordinates of the foci are \((h,k\pm c)\)
      • the equations of the asymptotes are \(y=\pm \dfrac{a}{b}(x−h)+k\)
  3. Solve for the coordinates of the foci using the equation \(c=\pm \sqrt{a^2+b^2}\).
  4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.

Example \(\PageIndex{5}\): Graphing a Hyperbola Centered at \((h, k)\) Given an Equation in General Form

Graph the hyperbola given by the equation \(9x^2−4y^2−36x−40y−388=0\). Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Solution

Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation.

\((9x^2−36x)−(4y^2+40y)=388\)

Factor the leading coefficient of each expression.

\(9(x^2−4x)−4(y^2+10y)=388\)

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

\(9(x^2−4x+4)−4(y^2+10y+25)=388+36−100\)

Rewrite as perfect squares.

\(9{(x−2)}^2−4{(y+5)}^2=324\)

Divide both sides by the constant term to place the equation in standard form.

\(\dfrac{{(x−2)}^2}{36}−\dfrac{{(y+5)}^2}{81}=1\)

The standard form that applies to the given equation is \(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\), where \(a^2=36\) and \(b^2=81\),or \(a=6\) and \(b=9\). Thus, the transverse axis is parallel to the \(x\)-axis. It follows that:

the center of the ellipse is \((h,k)=(2,−5)\)

the coordinates of the vertices are \((h\pm a,k)=(2\pm 6,−5)\), or \((−4,−5)\) and \((8,−5)\)

the coordinates of the co-vertices are \((h,k\pm b)=(2,−5\pm 9)\), or \((2,−14)\) and \((2,4)\)

the coordinates of the foci are \((h\pm c,k)\), where \(c=\pm \sqrt{a^2+b^2}\). Solving for \(c\),we have

\(c=\pm \sqrt{36+81}=\pm \sqrt{117}=\pm 3\sqrt{13}\)

Therefore, the coordinates of the foci are \((2−3\sqrt{13},−5)\) and \((2+3\sqrt{13},−5)\).

The equations of the asymptotes are \(y=\pm \dfrac{b}{a}(x−h)+k=\pm \dfrac{3}{2}(x−2)−5\).

Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola, as shown in Figure \(\PageIndex{10}\).

A horizontal hyperbola centered at (2, negative 5) with Vertices at (negative 4, negative 5) and (8, 5) and Foci at (2 minus 3 square root of 13, negative 5) and (2 + 3 square root of 13, negative 5). Also shown are the slant asymptotes, y = (3/2) times (x minus 2) minus 5 and y = (negative 3/2)times (x minus 2) minus 5. The points (2, negative 14), (2, 4) and (0, 0) are labeled.

Exercise \(\PageIndex{5}\)

Graph the hyperbola given by the standard form of an equation \(\dfrac{{(y+4)}^2}{100}−\dfrac{{(x−3)}^2}{64}=1\). Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Answer

center: \((3,−4)\); vertices: \((3,−14)\) and \((3,6)\); co-vertices: \((−5,−4)\); and \((11,−4)\); foci: \((3,−4−2\sqrt{41})\) and \((3,−4+2\sqrt{41})\); asymptotes: \(y=\pm \dfrac{5}{4}(x−3)−4\)

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Solving Applied Problems Involving Hyperbolas

As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength (Figure \(\PageIndex{12}\)). For example, a \(500\)-foot tower can be made of a reinforced concrete shell only \(6\) or \(8\) inches wide!

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The first hyperbolic towers were designed in 1914 and were \(35\) meters high. Today, the tallest cooling towers are in France, standing a remarkable \(170\) meters tall. In Example \(\PageIndex{6}\) we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides.

Example \(\PageIndex{6}\): Solving Applied Problems Involving Hyperbolas

The design layout of a cooling tower is shown in Figure \(\PageIndex{13}\). The tower stands \(179.6\) meters tall. The diameter of the top is \(72\) meters. At their closest, the sides of the tower are \(60\) meters apart.

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Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

Solution

We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin: \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\), where the branches of the hyperbola form the sides of the cooling tower. We must find the values of \(a^2\) and \(b^2\) to complete the model.

First, we find \(a^2\). Recall that the length of the transverse axis of a hyperbola is \(2a\). This length is represented by the distance where the sides are closest, which is given as \(65.3\) meters. So, \(2a=60\). Therefore, \(a=30\) and \(a^2=900\).

To solve for \(b^2\),we need to substitute for \(x\) and \(y\) in our equation using a known point. To do this, we can use the dimensions of the tower to find some point \((x,y)\) that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the \(y\)-axis bisects the tower, our \(x\)-value can be represented by the radius of the top, or \(36\) meters. The y-value is represented by the distance from the origin to the top, which is given as \(79.6\) meters. Therefore,

\[\begin{align*} \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}&=1\qquad \text{Standard form of horizontal hyperbola.}\\ b^2&=\dfrac{y^2}{\dfrac{x^2}{a^2}-1}\qquad \text{Isolate } b^2\\ &=\dfrac{{(79.6)}^2}{\dfrac{{(36)}^2}{900}-1}\qquad \text{Substitute for } a^2,\: x, \text{ and } y\\ &\approx 14400.3636\qquad \text{Round to four decimal places} \end{align*}\]

The sides of the tower can be modeled by the hyperbolic equation

\(\dfrac{x^2}{900}−\dfrac{y^2}{14400.3636}=1\),or \(\dfrac{x^2}{{30}^2}−\dfrac{y^2}{{120.0015}^2}=1\)

Exercise \(\PageIndex{6}\)

A design for a cooling tower project is shown in Figure \(\PageIndex{14}\). Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

Project design for a natural draft cooling tower. The overall height is 167.082 meters. The diameter at the top is 60 meters, and at their closest, 79.6 meters from the top, the sides are 60 meters apart.
Answer

The sides of the tower can be modeled by the hyperbolic equation. \(\dfrac{x^2}{400}−\dfrac{y^2}{3600}=1\) or \(\dfrac{x^2}{{20}^2}−\dfrac{y^2}{{60}^2}=1\).

Media

Access these online resources for additional instruction and practice with hyperbolas.

  • Conic Sections: The Hyperbola Part 1 of 2
  • Conic Sections: The Hyperbola Part 2 of 2
  • Graph a Hyperbola with Center at Origin
  • Graph a Hyperbola with Center not at Origin

Key Equations

Hyperbola, center at origin, transverse axis on x-axis\(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\)
Hyperbola, center at origin, transverse axis on y-axis\(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\)
Hyperbola, center at \((h,k)\),transverse axis parallel to x-axis\(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\)
Hyperbola, center at \((h,k)\),transverse axis parallel to y-axis\(\dfrac{{(y−k)}^2}{a^2}−\dfrac{{(x−h)}^2}{b^2}=1\)

Key Concepts

  • A hyperbola is the set of all points \((x,y)\) in a plane such that the difference of the distances between \((x,y)\) and the foci is a positive constant.
  • The standard form of a hyperbola can be used to locate its vertices and foci. See Example \(\PageIndex{1}\).
  • When given the coordinates of the foci and vertices of a hyperbola, we can write the equation of the hyperbola in standard form. See Example \(\PageIndex{2}\) and Example \(\PageIndex{3}\).
  • When given an equation for a hyperbola, we can identify its vertices, co-vertices, foci, asymptotes, and lengths and positions of the transverse and conjugate axes in order to graph the hyperbola. See Example \(\PageIndex{4}\) and Example \(\PageIndex{5}\).
  • Real-world situations can be modeled using the standard equations of hyperbolas. For instance, given the dimensions of a natural draft cooling tower, we can find a hyperbolic equation that models its sides. See Example \(\PageIndex{6}\).

Contributors and Attributions

Sours: https://math.libretexts.org/Bookshelves/Algebra/Map%3A_College_Algebra_(OpenStax)/08%3A_Analytic_Geometry/8.03%3A_The_Hyperbola

Foci formula hyperbola

Finding and Graphing the Foci of a Hyperbola

This hyperbola has already been graphed and its center point is marked:



We need to use the formula c2= a2+ b2to find c. Now, we could find a and b and then substitute, but remember that in the pattern, the denominators are a2and b2, so we can substitute those right into the formula:

c2 = a2 + b2

c2 = 9 + 16

c2 = 25 We'll need to take the square root.

c = 5

We can plot the foci by counting 5 spaces from the center.Note that we are notcounting 5 spaces from the vertex. Since the hyperbola is horizontal, we will count 5 spaces left and right and plot the foci there.



Sours: https://www.softschools.com/math/calculus/finding_and_graphing_the_foci_of_a_hyperbola/
Writing the equation of a hyperbola given the foci and vertices

Let [latex]\left(-c,0\right)[/latex] and [latex]\left(c,0\right)[/latex] be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points [latex]\left(x,y\right)[/latex] such that the difference of the distances from [latex]\left(x,y\right)[/latex] to the foci is constant.

Figure 4

If [latex]\left(a,0\right)[/latex] is a vertex of the hyperbola, the distance from [latex]\left(-c,0\right)[/latex] to [latex]\left(a,0\right)[/latex] is [latex]a-\left(-c\right)=a+c[/latex]. The distance from [latex]\left(c,0\right)[/latex] to [latex]\left(a,0\right)[/latex] is [latex]c-a[/latex]. The sum of the distances from the foci to the vertex is

[latex]\left(a+c\right)-\left(c-a\right)=2a[/latex]

If [latex]\left(x,y\right)[/latex] is a point on the hyperbola, we can define the following variables:

[latex]\begin{array}{l}{d}_{2}=\text{the distance from }\left(-c,0\right)\text{ to }\left(x,y\right)\\ {d}_{1}=\text{the distance from }\left(c,0\right)\text{ to }\left(x,y\right)\end{array}[/latex]

By definition of a hyperbola, [latex]{d}_{2}-{d}_{1}[/latex] is constant for any point [latex]\left(x,y\right)[/latex] on the hyperbola. We know that the difference of these distances is [latex]2a[/latex] for the vertex [latex]\left(a,0\right)[/latex]. It follows that [latex]{d}_{2}-{d}_{1}=2a[/latex] for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses.

[latex]\begin{array}{ll}\text{ }{d}_{2}-{d}_{1}=\sqrt{{\left(x-\left(-c\right)\right)}^{2}+{\left(y - 0\right)}^{2}}-\sqrt{{\left(x-c\right)}^{2}+{\left(y - 0\right)}^{2}}=2a\hfill & \text{Distance Formula}\hfill \\ \sqrt{{\left(x+c\right)}^{2}+{y}^{2}}-\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}=2a\hfill & \text{Simplify expressions}\text{.}\hfill \\ \text{ }\sqrt{{\left(x+c\right)}^{2}+{y}^{2}}=2a+\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill & \text{Move radical to opposite side}\text{.}\hfill \\ \text{ }{\left(x+c\right)}^{2}+{y}^{2}={\left(2a+\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\right)}^{2}\hfill & \text{Square both sides}\text{.}\hfill \\ \text{ }{x}^{2}+2cx+{c}^{2}+{y}^{2}=4{a}^{2}+4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}+{\left(x-c\right)}^{2}+{y}^{2}\hfill & \text{Expand the squares}\text{.}\hfill \\ \text{ }{x}^{2}+2cx+{c}^{2}+{y}^{2}=4{a}^{2}+4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}+{x}^{2}-2cx+{c}^{2}+{y}^{2}\hfill & \text{Expand remaining square}\text{.}\hfill \\ \text{ }2cx=4{a}^{2}+4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}-2cx\hfill & \text{Combine like terms}\text{.}\hfill \\ \text{ }4cx - 4{a}^{2}=4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill & \text{Isolate the radical}\text{.}\hfill \\ \text{ }cx-{a}^{2}=a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill & \text{Divide by 4}\text{.}\hfill \\ \text{ }{\left(cx-{a}^{2}\right)}^{2}={a}^{2}{\left[\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\right]}^{2}\hfill & \text{Square both sides}\text{.}\hfill \\ \text{ }{c}^{2}{x}^{2}-2{a}^{2}cx+{a}^{4}={a}^{2}\left({x}^{2}-2cx+{c}^{2}+{y}^{2}\right)\hfill & \text{Expand the squares}\text{.}\hfill \\ \text{ }{c}^{2}{x}^{2}-2{a}^{2}cx+{a}^{4}={a}^{2}{x}^{2}-2{a}^{2}cx+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\hfill & \text{Distribute }{a}^{2}\text{.}\hfill \\ \text{ }{a}^{4}+{c}^{2}{x}^{2}={a}^{2}{x}^{2}+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\hfill & \text{Combine like terms}\text{.}\hfill \\ \text{ }{c}^{2}{x}^{2}-{a}^{2}{x}^{2}-{a}^{2}{y}^{2}={a}^{2}{c}^{2}-{a}^{4}\hfill & \text{Rearrange terms}\text{.}\hfill \\ \text{ }{x}^{2}\left({c}^{2}-{a}^{2}\right)-{a}^{2}{y}^{2}={a}^{2}\left({c}^{2}-{a}^{2}\right)\hfill & \text{Factor common terms}\text{.}\hfill \\ \text{ }{x}^{2}{b}^{2}-{a}^{2}{y}^{2}={a}^{2}{b}^{2}\hfill & \text{Set }{b}^{2}={c}^{2}-{a}^{2}.\hfill \\ \text{ }\frac{{x}^{2}{b}^{2}}{{a}^{2}{b}^{2}}-\frac{{a}^{2}{y}^{2}}{{a}^{2}{b}^{2}}=\frac{{a}^{2}{b}^{2}}{{a}^{2}{b}^{2}}\hfill & \text{Divide both sides by }{a}^{2}{b}^{2}\hfill \\ \text{ }\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\hfill & \hfill \end{array}[/latex]

This equation defines a hyperbola centered at the origin with vertices [latex]\left(\pm a,0\right)[/latex] and co-vertices [latex]\left(0\pm b\right)[/latex].

A General Note: Standard Forms of the Equation of a Hyperbola with Center (0,0)

The standard form of the equation of a hyperbola with center [latex]\left(0,0\right)[/latex] and transverse axis on the x-axis is

[latex]\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1[/latex]

where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(\pm a,0\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(0,\pm b\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(\pm c,0\right)[/latex]
  • the equations of the asymptotes are [latex]y=\pm \frac{b}{a}x[/latex]

The standard form of the equation of a hyperbola with center [latex]\left(0,0\right)[/latex] and transverse axis on the y-axis is

[latex]\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1[/latex]

where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(0,\pm a\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(\pm b,0\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(0,\pm c\right)[/latex]
  • the equations of the asymptotes are [latex]y=\pm \frac{a}{b}x[/latex]

Note that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

Figure 5. (a) Horizontal hyperbola with center [latex]\left(0,0\right)[/latex] (b) Vertical hyperbola with center [latex]\left(0,0\right)[/latex]

How To: Given the equation of a hyperbola in standard form, locate its vertices and foci.

  1. Determine whether the transverse axis lies on the x– or y-axis. Notice that [latex]{a}^{2}[/latex] is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.

a. If the equation has the form [latex]\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1[/latex], then the transverse axis lies on the x-axis. The vertices are located at [latex]\left(\pm a,0\right)[/latex], and the foci are located at [latex]\left(\pm c,0\right)[/latex].

b. If the equation has the form [latex]\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1[/latex], then the transverse axis lies on the y-axis. The vertices are located at [latex]\left(0,\pm a\right)[/latex], and the foci are located at [latex]\left(0,\pm c\right)[/latex].

  1. Solve for [latex]a[/latex] using the equation [latex]a=\sqrt{{a}^{2}}[/latex].
  2. Solve for [latex]c[/latex] using the equation [latex]c=\sqrt{{a}^{2}+{b}^{2}}[/latex].

Example 1: Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation [latex]\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32}=1[/latex].

Solution

The equation has the form [latex]\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1[/latex], so the transverse axis lies on the y-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set [latex]x=0[/latex], and solve for [latex]y[/latex].

[latex]\begin{array}{l}1=\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32}\hfill \\ 1=\frac{{y}^{2}}{49}-\frac{{0}^{2}}{32}\hfill \\ 1=\frac{{y}^{2}}{49}\hfill \\ {y}^{2}=49\hfill \\ y=\pm \sqrt{49}=\pm 7\hfill \end{array}[/latex]

The foci are located at [latex]\left(0,\pm c\right)[/latex]. Solving for [latex]c[/latex],

[latex]c=\sqrt{{a}^{2}+{b}^{2}}=\sqrt{49+32}=\sqrt{81}=9[/latex]

Therefore, the vertices are located at [latex]\left(0,\pm 7\right)[/latex], and the foci are located at [latex]\left(0,9\right)[/latex].

Try It 1

Identify the vertices and foci of the hyperbola with equation [latex]\frac{{x}^{2}}{9}-\frac{{y}^{2}}{25}=1[/latex].

Solution

Sours: https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/deriving-the-equation-of-a-hyperbola-centered-at-the-origin/

Now discussing:

More on hyperbolas

A hyperbola is the set of all points in the plane such that the difference between the distances from to two fixed points is a given constant.  Each of the fixed points is a focus .  (The plural is foci.)  If is a point on the hyperbola and the foci are and then and are the focal radii .  As you can see, the graph of the hyperbola has two disconnected branches that look similar to parabolas.  Each branch approaches diagonal asymptotes .

   

The center of a hyperbola is the midpoint of the line segment joining its foci.  The transverse axis is the line segment that contains the center of the hyperbola and whose endpoints are the two vertices of the hyperbola.

 A central hyperbola, one with its center on the origin and its foci on either the -axis or -axis has one to the two formulas below.  Note that is the denominator of the positive term in each case.

 The hyperbola with its center at , foci and and distance between the vertices has the equation

where .

The equations of the asymptotes are and

Note that the hyperbola will have either - or -intercepts but never both.

   

The hyperbola with its center at , foci and , then the equation is with

The equations of the asymptotes are and .

Example 1:

Find an equation of the hyperbola having foci and and the distance between the vertices equal to if its center is at the origin.

   The distance from each focus to the center is , so . 

   The distance between the vertices is .  So, and .

, so

   Since the foci are on the -axis, the equation is .

Example 2:

For the above example, find the equations of its asymptotes and graph the hyperbola.

   The equations of the asymptotes are and .

   The -intercepts are and .  There are no -intercepts.

   

The graph of a hyperbola can be translated so that its center is at the point .  This means that the graph has been translated units on the horizontal axis and units on the vertical axis.

Horizontal Major Axis Vertical Major Axis

  

Foci at and Foci at and

Example 3:

Find an equation of the hyperbola with foci and and the distance between the vertices .

   The distance between the vertices, , is .  So, and .

   The center is halfway between the foci at .

   The distance from the center to each focus is so .

   Therefore, the equation of the hyperbola is

   

A hyperbola can also be defined as a conic section obtained by the intersection of a double cone with a plane that is intersects both pieces of the cone without intersecting the axis.

   

Sours: https://www.varsitytutors.com/hotmath/hotmath_help/topics/more-on-hyperbolas


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