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Answers without the blur. Sign up and see all textbooks for free! Q. 27

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. $\text{directrix}y={y}_{0}\text{, focus}\left({x}_{1},{y}_{1}\right)\text{, where}{y}_{0}\ne {y}_{1}$

The equation is $y=\frac{1}{2}·\frac{{\left(x-{x}_{1}\right)}^{2}}{\left({y}_{1}-{y}_{0}\right)}+\frac{\left({y}_{0}+{y}_{1}\right)}{2}$.

See the step by step solution

## Step 1. Given information.

The given values are,

$\text{directrix}y={y}_{0}\text{, focus}\left({x}_{1},{y}_{1}\right)\text{, where}{y}_{0}\ne {y}_{1}$

## Step 2. Distance formula.

Let$\left(x,y\right)$ be any point on the parabola.

Let $\left({x}_{1},{y}_{1}\right)$ be the focus.

Therefore, by distance formula,

$\text{Distance}=\sqrt{{\left({x}_{1}-x\right)}^{2}+{\left({y}_{1}-y\right)}^{2}}\left[\text{since}{x}_{1}=x,{y}_{1}=y,{x}_{2}={x}_{1},{y}_{2}={y}_{1}\right]\phantom{\rule{0ex}{0ex}}\text{Now the distance between the point and the directrix is}\left|y-{y}_{0}\right|\text{.}\phantom{\rule{0ex}{0ex}}Therefore,\phantom{\rule{0ex}{0ex}}\sqrt{{\left({x}_{1}-x\right)}^{2}+{\left({y}_{1}-y\right)}^{2}}=\left|y-{y}_{0}\right|$

On simplifying the equation,

${\left(\sqrt{{\left({x}_{1}-x\right)}^{3}+{\left({y}_{1}-y\right)}^{3}}\right)}^{2}={\left(\left|y-{y}_{0}\right|\right)}^{2}\phantom{\rule{0ex}{0ex}}{\left(x-{x}_{1}\right)}^{2}+{\left(y-{y}_{1}\right)}^{2}={\left(y-{y}_{0}\right)}^{2}\phantom{\rule{0ex}{0ex}}{\left(x-{x}_{1}\right)}^{2}=\left({y}^{2}+{y}_{0}^{2}-2y{y}_{0}\right)-\left({y}^{2}+{y}_{1}^{2}-2y{y}_{1}\right)\phantom{\rule{0ex}{0ex}}{\left(x-{x}_{1}\right)}^{2}=\left({y}_{0}^{2}-{y}_{1}^{2}\right)-2y\left({y}_{0}-{y}_{1}\right)\phantom{\rule{0ex}{0ex}}\frac{{\left(x-{x}_{1}\right)}^{2}}{\left({y}_{0}-{y}_{1}\right)}=\frac{\left({y}_{0}-{y}_{1}\right)\left(\left({y}_{0}+{y}_{1}\right)-2y\right)}{\left({y}_{0}-{y}_{1}\right)}\phantom{\rule{0ex}{0ex}}\frac{{\left(x-{x}_{1}\right)}^{2}}{\left({y}_{0}-{y}_{1}\right)}=\left({y}_{0}+{y}_{1}\right)-2y\phantom{\rule{0ex}{0ex}}or\phantom{\rule{0ex}{0ex}}-2y=\frac{{\left(x-{x}_{1}\right)}^{2}}{\left({y}_{0}-{y}_{1}\right)}-\left({y}_{0}+{y}_{1}\right)\phantom{\rule{0ex}{0ex}}\frac{-2y}{-2}=\frac{1}{-2}·\frac{{\left(x-{x}_{1}\right)}^{2}}{\left({y}_{0}-{y}_{1}\right)}-\frac{1}{-2}·\left({y}_{0}+{y}_{1}\right)\phantom{\rule{0ex}{0ex}}y=\frac{1}{2}·\frac{{\left(x-{x}_{1}\right)}^{2}}{\left({y}_{1}-{y}_{0}\right)}+\frac{\left({y}_{0}+{y}_{1}\right)}{2}$ ### Want to see more solutions like these? 