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Q. 22

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. $\text{directrix}x=3\text{, focus}\left(0,1\right)$

The equation is $\left(y-1{\right)}^{2}=-6x+9.$

See the step by step solution

## Step 1. Given information.

We are given,

$\text{directrix}x=3\text{, focus}\left(0,1\right)$

## Step 2. Distance formula.

Now,

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

We know,

$\text{Formula for the distance}=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\text{Distance}=\sqrt{\left(0-x{\right)}^{2}+\left(1-y{\right)}^{2}}\left[\text{since}{x}_{1}=x,{y}_{1}=y,{x}_{2}=0,{y}_{2}=1\right]$

Now, the distance between the point and directrix $\left|x-3\right|$

Then,

$\sqrt{\left(0-x{\right)}^{2}+\left(1-y{\right)}^{2}}=|x-3|$

${\left(\sqrt{\left(0-x{\right)}^{2}+\left(1-y{\right)}^{2}}\right)}^{2}=\left(|x-3|{\right)}^{2}\phantom{\rule{0ex}{0ex}}\left(0-x{\right)}^{2}+\left(1-y{\right)}^{2}=\left(x-3{\right)}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}+1+{y}^{2}-2y=\left({x}^{2}-6x+9\right)\phantom{\rule{0ex}{0ex}}1+{y}^{2}-2y=-6x+9\phantom{\rule{0ex}{0ex}}{y}^{2}-2y+1=-6x+9\phantom{\rule{0ex}{0ex}}\left(y-1{\right)}^{2}=-6x+9$