Chapter 9.7, Problem 52E

To determine

To calculate:

The polar equation of the conic by using the following data:

Conic section: Hyperbola

Vertices: $\left(2,0\right),\left(-8,\pi \right)$

Answer to Problem 52E

The polar equation of the hyperbola is derived as $r=\frac{\left(\frac{8}{3}\right)}{1-\frac{5}{3}\cos \theta }$

Explanation of Solution

Given information:

Conic section: Hyperbola

Vertices: $\left(2,0\right),\left(-8,\pi \right)$

Formula used:

The equation of the hyperbola is given by the following expression:

  $r=\frac{ep}{1\pm e\cos \theta }$

Calculation:

Here, the given conic section is hyperbola and its vertices are $\left(2,0\right),\left(-8,\pi \right)$ . Now need to find the equation of hyperbola:

The vertices $\left(2,0\right),\left(-8,\pi \right)$ lies on the $y-$ axis. Thus, on line of $\left(\theta =0\right)$ , the transverse axis of the hyperbola lies.

So the equation of the hyperbola is given by the following expression:

  $r=\frac{ep}{1+e\cos \theta }$

At vertex $\left(r,\theta \right)=\left(2,0\right)$ :

  $\begin{array}{l}r=\frac{ep}{1+e\cos \theta }\\ 2=\frac{ep}{1+e\cos \left(0\right)}\\ 2=\frac{ep}{1+e}\left[\because \cos \left(0\right)=1\right]\end{array}$

  $ep-2e-2=0$.........(1)

At vertex $\left(r,\theta \right)=\left(-8,\pi \right)$ :

  $\begin{array}{l}r=\frac{ep}{1+e\cos \theta }\\ -8=\frac{ep}{1+e\cos \left(\pi \right)}\\ -8=\frac{ep}{1-e}\left[\because \cos \left(\pi \right)=-1\right]\end{array}$

  $ep-8e+8=0$.........(2)

Take subtraction of eq. (1) and (2) to get value of $e$ :

  $\begin{array}{l}ep-2e-2-ep+8e-8=0\\ 6e=10\\ e=-\frac{5}{3}\end{array}$

From eq. (1):

  $\begin{array}{l}ep+e-1=0\\ -\frac{5}{3}p-\frac{5}{3}-1=0\\ -\frac{5}{3}p=\frac{8}{3}\\ P=-\frac{8}{5}\end{array}$

Hence, the equation of the hyperbola is derived as:

  $r=\frac{ep}{1+e\cos \theta }$

Put the calculated values:

  $\begin{array}{l}r=\frac{ep}{1+e\cos \theta }\\ r=\frac{\left(-\frac{5}{3}\right)\left(-\frac{8}{5}\right)}{1-\frac{5}{3}\cos \theta }\\ r=\frac{\left(\frac{8}{3}\right)}{1-\frac{5}{3}\cos \theta }\end{array}$