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#?

Question

1900 views around the world

You can reuse this answer
Creative Commons License

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)#].

,