How do you find the equation in standard form of the parabola with directrix x=5 and focus (11, -7)?

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Answer 1 · 2018-07-21T14:41:58

The equation is $= x = \frac{1}{12} {(y + 7)}^{2} + 8$ $=x=1/12(y+7)^2+8$

Any point $(x, y)$ $(x,y)$ on the parabola is equidistant from the directrix and from the focus

$(x - 5) = \sqrt{{(x - 11)}^{2} + {(y + 7)}^{2}}$ $(x-5)=sqrt((x-11)^2+(y+7)^2)$

Squaring both sides

${(x - 5)}^{2} = {(x - 11)}^{2} + {(y + 7)}^{2}$ $(x-5)^2=(x-11)^2+(y+7)^2$

$x^{2} - 10 x + 25 = x^{2} - 22 x + 121 + {(y + 7)}^{2}$ $x^2-10x+25=x^2-22x+121+(y+7)^2$

$12 x = {(y + 7)}^{2} + 96$ $12x=(y+7)^2+96$

$x = \frac{1}{12} {(y + 7)}^{2} + 8$ $x=1/12(y+7)^2+8$

graph{(x-1/12(y+7)^2-8)=0 [-14.88, 50.06, -18.7, 13.76]}