Chapter 9.7, Problem 53E

To determine

To calculate:

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

Conic section: Hyperbola

Vertices: $\left(\frac{3}{5},\pi \right),\left(-1,0\right)$

Answer to Problem 53E

The polar equation of the hyperbola is derived as $r=\frac{3}{1-4\cos \theta }$ .

Explanation of Solution

Given information:

Conic section: Hyperbola

Vertices: $\left(\frac{3}{5},\pi \right),\left(-1,0\right)$

Formula used:

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

  $r=\frac{ep}{1\pm e\sin \theta }$

Calculation:

Here, the given conic section is hyperbola and its vertices are $\left(\frac{3}{5},\pi \right),\left(-1,0\right)$ . Now need to find the equation of hyperbola:

The vertices $\left(2,0\right),\left(-8,\pi \right)$ lies on the $x-$ 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 $\left(r,\theta \right)=\left(\frac{3}{5},\pi \right)$ :

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

  $5ep+3e-3=0$.........(1)

At vertex $\left(r,\theta \right)=\left(-1,0\right)$ :

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

  $ep+e+1=0$.........(2)

Multiply by $5$ :

  $5ep+5e+5=0$.........(3)

Take subtraction of eq. (1) and (3) to get value of $e$ :

  $\begin{array}{l}5ep+3e-3-5ep-5e-5=0\\ -2e=8\\ e=-4\end{array}$

From eq. (2):

  $\begin{array}{l}ep+e+1=0\\ -4p-4+1=0\\ p=-\frac{3}{4}\end{array}$

Hence, the equation of the hyperbola is derived as:

  $r=\frac{ep}{1+e\cos \theta }$

Put the calculated values:

  $\begin{array}{l}r=\frac{ep}{1+e\cos \theta }\\ r=\frac{\left(-4\right)\left(-\frac{3}{4}\right)}{1-4\cos \theta }\\ r=\frac{3}{1-4\cos \theta }\end{array}$