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

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Answer 1 · 2016-09-01T14:31:59

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

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