# How do you find the standard form of the equation of the hyperbola given the properties vertices (3,2), (13,2); endpoints of the conjugate axis (8,4), (8,0)?

##### 1 Answer
May 17, 2017

Equation of hyperbola is ${\left(x - 8\right)}^{2} / 25 - {\left(y - 2\right)}^{2} / 4 = 1$

#### Explanation:

As the ordinates of vertices are equal, the equation of hyperbola is of the form ${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$, where $\left(h , k\right)$ is the centre of hyperbola. As vertices are $\left(3 , 2\right)$ and $\left(13 , 2\right)$, centre is $\left(\frac{3 + 13}{2} , \frac{2 + 2}{2}\right)$ i.e. $\left(8 , 2\right)$ and hence equation of hyperbola is

${\left(x - 8\right)}^{2} / {a}^{2} - {\left(y - 2\right)}^{2} / {b}^{2} = 1$

To find $a$ and $b$, we will work them out using their relation with vertices and endpoints of conjugate axis. Now length of conjugate axis is $2 b$ and as the distance between endpoints is $4$, we have $2 b = 4$ i.e. $b = 2$.

We have the distance between vertices as $10$ and as this is $2 a$, we have $a = 5$ and hence equation of hyperbola is

${\left(x - 8\right)}^{2} / 25 - {\left(y - 2\right)}^{2} / 4 = 1$

graph{((x-8)^2/25-(y-2)^2/4-1)((x-13)^2+(y-2)^2-0.03)((x-3)^2+(y-2)^2-0.03)=0 [-12, 28, -7.92, 12.08]}