What is the standard form of the equation of the parabola with a directrix at x=-3 and a focus at (6,2)?

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Answer 1 · 2017-12-31T11:56:57

The standard equation of horizontal parabola is
${(y - 2)}^{2} = 18 (x - 1.5)$ $(y-2)^2 = 18(x-1.5)$

Focus is at $(6, 2)$ $(6,2)$ and directrix is $x = - 3$ $x=-3$ . Vertex is at midway

between focus and directrix. Therefore vertex is at

$(\frac{6 - 3}{2}, 2) or (1.5, 2)$ $((6-3)/2,2) or (1.5,2)$ .Here the directrix is at left of

the vertex , so parabola opens right and $p$ $p$ is positive.

The standard equation of horizontal parabola opening right is

${(y - k)}^{2} = 4 p (x - h); h = 1.5, k = 2$ $(y-k)^2 = 4p(x-h) ; h=1.5 ,k=2$

or ${(y - 2)}^{2} = 4 p (x - 1.5)$ $(y-2)^2 = 4p(x-1.5)$ The distance between focus and

vertex is $p = 6 - 1.5 = 4.5$ $p=6-1.5=4.5$ . Thus the standard equation of

horizontal parabola is ${(y - 2)}^{2} = 4 \cdot 4.5 (x - 1.5)$ $(y-2)^2 = 4*4.5(x-1.5)$ or

${(y - 2)}^{2} = 18 (x - 1.5)$ $(y-2)^2 = 18(x-1.5)$

graph{(y-2)^2=18(x-1.5) [-40, 40, -20, 20]}