Q. 29 - Calculus

Question

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

$r=\frac{4}{1+\sin \theta }$

Step-by-Step Solution

Answer

The equation is $y=-\frac{1}{8}{x}^2+2.$

Step 1. Given information.

The given equation in polar form is $r=\frac{4}{1+\sin \theta }$.

Step 2. Conversion.

On converting the polar-co-ordinates into cartesian co-ordinates,

$$\begin{gathered} r=\frac{4}{1+\sin \theta } \\ r+r\sin \theta =4 \\ \sqrt{x^2+y^2}+r·\frac{y}{r}=4 \left[\text{ since }r=\sqrt{x^2+y^2}\text{ and }r\sin \theta =y\right] \\ \sqrt{x^2+y^2}+y=4 \end{gathered}$$

Step 3. Final answer.

On simplifying the equation,

$$\begin{gathered} \sqrt{x^2+y^2}+y-y=4-y \\ \sqrt{x^2+y^2}=4-y \\ {\left(\sqrt{x^2+y^2}\right)}^2=(4-y{)}^2 \\ {x}^2+y^2=16+y^2-8y \\ {x}^2+y^2-y^2-16=16+y^2-8y-y^2-16 \\ {x}^2-16=-8y \\ \frac{-8y}{-8}=\frac{x^2-16}{-8} \\ y=-\frac{1}{8}{x}^2+2 \end{gathered}$$