Chapter 10, Problem 42E

Solution

To determine : The equation of a hyperbola with vertices $\left(3,3\right)\text{ and }\left(9,3\right)$ , and foci $\left(1,3\right)\text{ and }\left(11,3\right)$ .

The required equation is $\frac{{\left(x-6\right)}^2}{9}-\frac{{\left(y-3\right)}^2}{16}=1$ .

Given information :

The verticesof the hyperbola is $\left(3,3\right)\text{ and }\left(9,3\right)$ , and foci are $\left(1,3\right)\text{ and }\left(11,3\right)$ .

Formula used:

The equation of anellipse with center $\left(h,k\right)$ and has horizontal transverse axis is given by:

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

Explanation :

Given vertices are $\left(3,3\right)\text{ and }\left(9,3\right)$ and foci are $\left(1,3\right)\text{ and }\left(11,3\right)$ .

Here, $y-$ coordinate of vertices are constant and is equal to $3$ .

So, transverse axis is a horizontal axis $y=3$ .

Distance between the vertices is

  $\begin{array}{l}2a=6\\ a=3\end{array}$

Centre is the midpoint of two vertices.

  $\begin{array}{l}\left(\frac{3+9}{2},\frac{3+3}{2}\right)\\ =\left(6,3\right)\end{array}$

So, the centre is $\left(h,k\right)=\left(6,3\right)$

To find $b$ , use the formula $c^2=a^2+b^2$ .

  $\begin{array}{l}{5}^2=3^2+b^2\\ 25=9+b^2\\ b^2=16\\ b=4\end{array}$

The equation of an ellipse with center $\left(h,k\right)$ and has horizontal transverse axis is given by:

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

Substitute $a=3,b=4,h=6\text{ and }k=3$ in the equation $\left(1\right)$ .

  $\begin{array}{l}\frac{{\left(x-6\right)}^2}{3^2}-\frac{{\left(y-3\right)}^2}{4^2}=1\\ \frac{{\left(x-6\right)}^2}{9}-\frac{{\left(y-3\right)}^2}{16}=1\end{array}$

Therefore, the required equation is $\frac{{\left(x-6\right)}^2}{9}-\frac{{\left(y-3\right)}^2}{16}=1$ .