Q. 28 - Calculus

Question

Use Cartesian coordinates to express the equations for the parabolas determined by the conditions specified in Exercises 22–31. 

$r=\frac{3}{1+\cos \theta }$

Step-by-Step Solution

Answer

The equation is $y^2=9-6x.$

Step 1. Given information.

The give equation is $r=\frac{3}{1+\cos \theta }.$

Step 2. Conversion.

Converting the polar co-ordinates into cartesian co-ordinates,

$$\begin{gathered} r=\frac{3}{1+\cos \theta } \\ r(1+\cos \theta )=3 \\ r+r\cos \theta =3 \\ \sqrt{x^2+y^2}+r·\frac{x}{r}=3\left[\text{ since }r=\sqrt{x^2+y^2}\text{ and }r\cos \theta =x\right] \\ \sqrt{x^2+y^2}+x=3 \end{gathered}$$

Step 3. Final answer.

On simplifying the equation,

$$\begin{gathered} \sqrt{x^2+y^2}+x-x=3-x \\ \sqrt{x^2+y^2}=3-x \\ {\left(\sqrt{x^2+y^2}\right)}^2=(3-x{)}^2 \\ {x}^2+y^2=9+x^2-6x \\ {x}^2+y^2-x^2=9+x^2-6x-x^2 \\ {y}^2=9-6x \end{gathered}$$