What is the equation of the parabola with a focus at (3,6) and a directrix of y= 8?

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What is the equation of the parabola with a focus at (3,6) and a directrix of y= 8?

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Answer 1 · 2016-02-14T19:14:05

$y = (- \frac{1}{4}) x^{2} + (\frac{6}{4}) x + (\frac{19}{4})$ $y=(-1/4)x^2+(6/4)x+(19/4)$

Explanation:

If the focus of a parabola is (3,6) and the directrix is y = 8, find the equation of the parabola.

Let ( x0 , y0 ) be any point on the parabola. First of all, finding the distance between (x0 , y0) and the focus. Then finding the distance between (x0 , y0) and directrix. Equating these two distance equations and the simplified equation in x0 and y0 is equation of the parabola.

The distance between (x0 , y0) and (3,6) is
$\sqrt{{(x 0 - 2)}^{2} + {(y 0 - 5)}^{2}}$ $sqrt((x0-2)^2+(y0-5)^2$

The distance between (x0 , y0) and the directrix, y = 8 is | y0– 8|.

Equating the two distance expressions and square on both sides.
$\sqrt{{(x 0 - 3)}^{2} + {(y 0 - 6)}^{2}}$ $sqrt((x0-3)^2+(y0-6)^2$ = | y0– 8|.

${(x 0 - 3)}^{2} + {(y 0 - 6)}^{2}$ $(x0-3)^2+(y0-6)^2$ = ${(y 0 - 8)}^{2}$ $(y0-8)^2$

Simplifying and bringing all terms to one side:

$x 0^{2} - 6 x 0 + 4 y 0 - 19 = 0$ $x0^2-6x0+4y0-19=0$

Write the equation with y0 on one side:
$y 0 = (- \frac{1}{4}) x 0^{2} + (\frac{6}{4}) x 0 + (\frac{19}{4})$ $y0=(-1/4)x0^2+(6/4)x0+(19/4)$

This equation in (x0 , y0) is true for all other values on the parabola and hence we can rewrite with (x , y).

So, the equation of the parabola with focus (3,6) and directrix is y = 8 is
$y = (- \frac{1}{4}) x^{2} + (\frac{6}{4}) x + (\frac{19}{4})$ $y=(-1/4)x^2+(6/4)x+(19/4)$

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What is the equation of the parabola with a focus at (3,6) and a directrix of y= 8?

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