Question

How do you write the equation for a hyperbola given vertices $\left( -5,0 \right)$ and $\left( 5,0 \right)$, conjugate axis of length $12$ units?

Answer 1

In this problem we need to calculate the equation of the hyperbola with a given vertex and conjugate axis length. On observing the vertices, we have the same $y$ coordinates, so the hyperbola is horizontal transverse axis type. For this we will assume the equation of the hyperbola as $\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$. From this equation we can write the coordinates of the vertices and equate them with the given values to get the values of $h$, $k$, $a$. Now we will equate the length of the conjugate axis to $2b$ and simplify the equation to get the value of $b$. Now we will substitute all the values we have in the equation of hyperbola to get the required result.

Complete step by step answer:
Given that, $\left( -5,0 \right)$ and $\left( 5,0 \right)$ are the vertices of the hyperbola and the length of the conjugate axis is $12$ units.
We can observe that the $y$ coordinates of the given vertices are equal. So, the hyperbola is a horizontal transverse axis type. Hence assume the equation of the hyperbola as
$\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$.
For the above hyperbola the coordinates of vertices will be $\left( h+a,k \right)$, $\left( h-a,k \right)$. Comparing these values with the values of given vertices, then we will get
$k=0$, $h+a=5$, $h-a=-5$.
Solving the equations $h+a=5$, $h-a=-5$, then we will get
$h=0$, $a=5$.
Now the length of the conjugate axis of the hyperbola is $2b$. Equating this value with the given conjugate axis length then we will have
$\begin{aligned} & \Rightarrow 2b=12 \\\ & \Rightarrow b=6 \\\ \end{aligned}$
Substituting the all the values we have in the equation of the hyperbola, then we will get
$\begin{aligned} & \Rightarrow \dfrac{{{\left( x-0 \right)}^{2}}}{{{5}^{2}}}-\dfrac{{{\left( y-0 \right)}^{2}}}{{{6}^{2}}}=1 \\\ & \Rightarrow \dfrac{{{x}^{2}}}{{{5}^{2}}}-\dfrac{{{y}^{2}}}{{{6}^{2}}}=1 \\\ \end{aligned}$
Hence the equation of the required hyperbola is $\dfrac{{{x}^{2}}}{25}-\dfrac{{{y}^{2}}}{36}=1$ and the diagram of the hyperbola is given by

Note:
In this problem we have the $y$ coordinates of the vertices are same so we have assumed the equation of the hyperbola as $\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$. If we have the $x$ coordinates of the vertices are same then we need to assume the equation of the hyperbola as $\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}-\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}=1$.