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Q. 22

Expert-verifiedFound in: Page 772

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.

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

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

We are given,

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

Now,

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

We know,

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

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

Then,

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

On simplifying the equation,

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

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