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Q. 31

Expert-verified
Found in: Page 772

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Use Cartesian coordinates to express the equations for the parabolas determined by the conditions specified in Exercises 22–31. $r=\frac{\alpha }{1+\mathrm{sin}\theta }$

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

See the step by step solution

## Step 1. Given information.

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

## Step 2. Conversion.

Converting the polar into cartesian co-ordinates,

$r=\frac{\alpha }{1+\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}r\left(1+\mathrm{sin}\theta \right)=\alpha \phantom{\rule{0ex}{0ex}}r+r\mathrm{sin}\theta =\alpha \phantom{\rule{0ex}{0ex}}\sqrt{{x}^{2}+{y}^{2}}+r·\frac{y}{r}=\alpha \left[\text{since}r=\sqrt{{x}^{2}+{y}^{2}}\text{and}r\mathrm{sin}\theta =y\right]\phantom{\rule{0ex}{0ex}}\sqrt{{x}^{2}+{y}^{2}}+y=\alpha$

$\sqrt{{x}^{2}+{y}^{2}}+\\mathrm{not}y-\\mathrm{not}y=\alpha -y\phantom{\rule{0ex}{0ex}}\sqrt{{x}^{2}+{y}^{2}}=\alpha -y\phantom{\rule{0ex}{0ex}}{\left(\sqrt{{x}^{2}+{y}^{2}}\right)}^{2}=\left(\alpha -y{\right)}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}+{y}^{2}={\alpha }^{2}+{y}^{2}-2y\alpha \phantom{\rule{0ex}{0ex}}{x}^{2}+{y}^{2}-{y}^{2}={\alpha }^{2}+{y}^{2}-2y\alpha -{y}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}=-2y\alpha +{\alpha }^{2}$