Chapter 9.7, Problem 75E

To determine

To calculate:

Polar form of the equation of the conic.

Answer to Problem 75E

Polar form of the equation of the conic is $\frac{144}{25co{s}^2\theta -16}$

Explanation of Solution

Given information:

Hyperbola:

One focus: $\left(5,0\right)$

Vertices: $\left(4,0\right)\left(4,\pi \right)$

Calculation:

As it is given that vertices are,\

  $\left(4,\frac{\pi }{2}\right)\left(4,-\frac{\pi }{2}\right)$

Therefore, the length of the transverse axis,

  $\begin{array}{l}2a=4+4\\ 2a=8\\ a=4\end{array}$

The focus is

  $\left(5,\frac{\pi }{2}\right)$

Since the center is midpoint thus center is,

  $\begin{array}{l}=\left(\frac{4-4}{2},\frac{\pi }{2}\right)\\ =\left(0,\frac{\pi }{2}\right)\end{array}$

Therefore, the distance of the center from focus

  $\begin{array}{l}c=5-0\\ =5\end{array}$

Therefore, eccentricity

Thus,

  $\begin{array}{l}{b}^2=a^2\left(e^2-1\right)\\ =4^2\left({\left(\frac{5}{4}\right)}^2-1\right)\\ =16\left(\frac{9}{16}\right)\\ =9\end{array}$

Therefore, the equation of the hyperbola is

  $\begin{array}{l}{r}^2=\frac{-b^2}{1-e^{2}co{s}^2\theta }\\ =\frac{-9^2}{1-{\left(\frac{5}{4}\right)}^{2}co{s}^2\theta }\\ =\frac{-9}{1-\left(\frac{25}{16}\right)co{s}^2\theta }\\ =\frac{144}{25co{s}^2\theta -16}\end{array}$