Chapter 9, Problem 49CLT

To determine

Find the equation in the rectangular form.

Answer to Problem 49CLT

The equation in the rectangular form is $9x^2-16y^2+20x+4=0$ .

Explanation of Solution

Given:

It is given in the question that the polar equation is $r=\frac{2}{4-5\cos \theta }$ .

Concept Used:

In this, use the concept of the relationship $x^2+y^2=r^2;x=r\cos \theta ,y=r\sin \theta$ , $\tan \theta =\frac{y}{x}$ .

Calculation:

Here, the equation is $r=\frac{2}{4-5\cos \theta }$ .

It has to be use the relationship and put the value of x and y in the equation:

  $\begin{array}{l}{x}^2+y^2=r^2\\ x=r\cos \theta \\ y=r\sin \theta \end{array}$

  $\tan \theta =\frac{y}{x}$

Substitute $\cos \theta =\frac{x}{r}$ to the polar equation:

  $\begin{array}{l}r=\frac{2}{4-5\cos \theta }\\ r=\frac{2}{4-5(\frac{x}{r})}\\ r=\frac{2}{4-5(\frac{x}{r})}\cdot \frac{x}{r}(Multiplytherightsideby\frac{x}{r})\\ r=\frac{2r}{4r-5x}\\ 4r-5x=2(Crossmultiply,cancellingr)\\ 4r=5x+2(Add5xtobothsides)\\ 16r^2=25x^2+20x+4(squarebothsides)\\ 16(x^2+y^2)=25x^2+20x+4(Use{x}^2+y^2=r^2)\end{array}$

Expand and simplify on one side:

  $\begin{array}{l}16x^2+16y^2=25x^2+20x+4\\ 9x^2-16y^2+20x+4=0\end{array}$

Conclusion:

The equation is $9x^2-16y^2+20x+4=0$ .