What is the vertex form of the equation of the parabola with a focus at (11,28) and a directrix of #y=21 #?

1 Answer
Binayaka C.
Jul 18, 2016

The equation of parabola in vertex form is #y=1/14(x-11)^2+24.5#

Explanation:

The Vertex is equuidistant from focus(11,28) and directrix (y=21). So vertex is at #11,(21+7/2)=(11,24.5)#
The equation of parabola in vertex form is #y=a(x-11)^2+24.5#. The distance of vertex from directrix is #d=24.5-21=3.5# We know, #d=1/(4|a|) or a=1/(4*3.5)=1/14#.Since Parabola opens up, 'a' is +ive.
Hence the equation of parabola in vertex form is #y=1/14(x-11)^2+24.5# graph{1/14(x-11)^2+24.5 [-160, 160, -80, 80]}[Ans]

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