What is the vertex form of the equation of the parabola with a focus at (8,7) and a directrix of $y = 18$ $y=18$ ?

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Answer 1 · 2016-05-30T13:14:36

$y = - \frac{1}{22} {(x - 8)}^{2} + \frac{25}{2}$ $y=-1/22(x-8)^2+25/2$

Let their be a point $(x, y)$ $(x,y)$ on parabola. Its distance from focus at $(8, 7)$ $(8,7)$ is

$\sqrt{{(x - 8)}^{2} + {(y - 7)}^{2}}$ $sqrt((x-8)^2+(y-7)^2)$

and its distance from directrix $y = 18$ $y=18$ will be $| y - 18 |$ $|y-18|$

Hence equation would be

$\sqrt{{(x - 8)}^{2} + {(y - 7)}^{2}} = (y - 18)$ $sqrt((x-8)^2+(y-7)^2)=(y-18)$ or

${(x - 8)}^{2} + {(y - 7)}^{2} = {(y - 18)}^{2}$ $(x-8)^2+(y-7)^2=(y-18)^2$ or

$x^{2} - 16 x + 64 + y^{2} - 14 y + 49 = y^{2} - 36 y + 324$ $x^2-16x+64+y^2-14y+49=y^2-36y+324$ or

$x^{2} - 16 x + 22 y - 211 = 0$ $x^2-16x+22y-211=0$

or $22 y = - x^{2} + 16 x + 211$ $22y=-x^2+16x+211$

or $y = - \frac{1}{22} (x^{2} - 16 x + 64) + \frac{211}{22} + \frac{64}{22}$ $y=-1/22(x^2-16x+64)+211/22+64/22$

or $y = - \frac{1}{22} {(x - 8)}^{2} + \frac{275}{22}$ $y=-1/22(x-8)^2+275/22$

or $y = - \frac{1}{22} {(x - 8)}^{2} + \frac{25}{2}$ $y=-1/22(x-8)^2+25/2$

graph{y=-1/22(x-8)^2+25/2 [-31.84, 48.16, -12.16, 27.84]}

What is the vertex form of the equation of the parabola with a focus at (8,7) and a directrix of y=18y=18 ?