Q. 30 - Calculus

Question

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

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

Step-by-Step Solution

Answer

The equation is $y^2=-2x\alpha +{\alpha }^2.$$y^2=-2x\alpha +{\alpha }^2.$

Step 1. Given information.

The given polar equation is $r=\frac{\alpha }{1+\cos \theta }.$

Step 2. Conversion.

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

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

Step 3. Final answer.

On simplifying the equation,

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