Question

How do you write the equation of a parabola in vertex form that has a focus at `(6,5)` and a directrix of `y=-1`?

Answer 1

`(x-6)^2=12(y-2)," is the reqd. eqn. of Parabola, with, Vertex "(6,2)`. graph{(x-6)^2=12(y-2) [-10, 10, -5, 5]}

Explanation:

Let the Focus be `S(6,5)` and, the Directrix `d : y+1=0`.

Let `P(x,y)` is any pt. on the reqd. Parabola, then, by the Focus-

Directrix Property of Parabola, we have,

`"Dist. "SP="the" bot-dist. "btwn. P & d"`.

`:. sqrt{(x-6)^2+(y-5)^2}=|y+1|`.

`:. (x-6)^2+(y-5)^2=(y+1)^2`.

`:. (x-6)^2=2y+1+10y-25=12y-24`, i.e.,

`(x-6)^2=12(y-2),` is the reqd. eqn. of Parabola [ Vertex `(6,2)`].

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