# 12.2 The hyperbola  (Page 4/13)

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## 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 $\text{\hspace{0.17em}}\left(±6,0\right)\text{\hspace{0.17em}}$ and foci $\text{\hspace{0.17em}}\left(±2\sqrt{10},0\right)?$

The vertices and foci are on the x -axis. Thus, the equation for the hyperbola will have the form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1.$

The vertices are $\text{\hspace{0.17em}}\left(±6,0\right),\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}a=6\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{a}^{2}=36.$

The foci are $\text{\hspace{0.17em}}\left(±2\sqrt{10},0\right),\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}c=2\sqrt{10}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{c}^{2}=40.$

Solving for $\text{\hspace{0.17em}}{b}^{2},$ we have

Finally, we substitute $\text{\hspace{0.17em}}{a}^{2}=36\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{b}^{2}=4\text{\hspace{0.17em}}$ into the standard form of the equation, $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1.\text{\hspace{0.17em}}$ The equation of the hyperbola is $\text{\hspace{0.17em}}\frac{{x}^{2}}{36}-\frac{{y}^{2}}{4}=1,$ as shown in [link] .

What is the standard form equation of the hyperbola that has vertices $\text{\hspace{0.17em}}\left(0,±2\right)\text{\hspace{0.17em}}$ and foci $\text{\hspace{0.17em}}\left(0,±2\sqrt{5}\right)?$

$\frac{{y}^{2}}{4}-\frac{{x}^{2}}{16}=1$

## 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 $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ units horizontally and $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ units vertically, the center of the hyperbola    will be $\text{\hspace{0.17em}}\left(h,k\right).\text{\hspace{0.17em}}$ This translation results in the standard form of the equation we saw previously, with $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ replaced by $\text{\hspace{0.17em}}\left(x-h\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ replaced by $\text{\hspace{0.17em}}\left(y-k\right).$

## Standard forms of the equation of a hyperbola with center ( h , k )

The standard form of the equation of a hyperbola with center $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ and transverse axis parallel to the x -axis is

$\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1\text{\hspace{0.17em}}$

where

• the length of the transverse axis is $\text{\hspace{0.17em}}2a$
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(h±a,k\right)$
• the length of the conjugate axis is $\text{\hspace{0.17em}}2b$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(h,k±b\right)$
• the distance between the foci is $\text{\hspace{0.17em}}2c,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}{c}^{2}={a}^{2}+{b}^{2}$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(h±c,k\right)$

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is $\text{\hspace{0.17em}}2a\text{\hspace{0.17em}}$ and its width is $\text{\hspace{0.17em}}2b.\text{\hspace{0.17em}}$ The slopes of the diagonals are $\text{\hspace{0.17em}}±\frac{b}{a},$ and each diagonal passes through the center $\text{\hspace{0.17em}}\left(h,k\right).\text{\hspace{0.17em}}$ Using the point-slope formula , it is simple to show that the equations of the asymptotes are $\text{\hspace{0.17em}}y=±\frac{b}{a}\left(x-h\right)+k.\text{\hspace{0.17em}}$ See [link] a

The standard form of the equation of a hyperbola with center $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ and transverse axis parallel to the y -axis is

$\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1$

where

• the length of the transverse axis is $\text{\hspace{0.17em}}2a$
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(h,k±a\right)$
• the length of the conjugate axis is $\text{\hspace{0.17em}}2b$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(h±b,k\right)$
• the distance between the foci is $\text{\hspace{0.17em}}2c,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}{c}^{2}={a}^{2}+{b}^{2}$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(h,k±c\right)$

Using the reasoning above, the equations of the asymptotes are $\text{\hspace{0.17em}}y=±\frac{a}{b}\left(x-h\right)+k.\text{\hspace{0.17em}}$ See [link] b .

Like hyperbolas centered at the origin, hyperbolas centered at a point $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ have vertices, co-vertices, and foci that are related by the equation $\text{\hspace{0.17em}}{c}^{2}={a}^{2}+{b}^{2}.\text{\hspace{0.17em}}$ 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.

Given the vertices and foci of a hyperbola centered at $\text{\hspace{0.17em}}\left(h,k\right),$ 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 $\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1.$
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 $\text{\hspace{0.17em}}\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1.$
2. Identify the center of the hyperbola, $\text{\hspace{0.17em}}\left(h,k\right),$ using the midpoint formula and the given coordinates for the vertices.
3. Find $\text{\hspace{0.17em}}{a}^{2}\text{\hspace{0.17em}}$ by solving for the length of the transverse axis, $\text{\hspace{0.17em}}2a$ , which is the distance between the given vertices.
4. Find $\text{\hspace{0.17em}}{c}^{2}\text{\hspace{0.17em}}$ using $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ found in Step 2 along with the given coordinates for the foci.
5. Solve for $\text{\hspace{0.17em}}{b}^{2}\text{\hspace{0.17em}}$ using the equation $\text{\hspace{0.17em}}{b}^{2}={c}^{2}-{a}^{2}.$
6. Substitute the values for $\text{\hspace{0.17em}}h,k,{a}^{2},$ and $\text{\hspace{0.17em}}{b}^{2}\text{\hspace{0.17em}}$ into the standard form of the equation determined in Step 1.

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