What is the vertex form of the equation of the parabola with a focus at (6,5) and a directrix of #y=19 #?

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Answer 1 · 2016-05-22T12:14:16

Vertex form of the equation of the parabola is
#y=-1/28(x-6)^2+12#

Here the directrix is a horizontal line #y=19#.

Since this line is perpendicular to the axis of symmetry, this is a regular parabola, where the #x# part is squared.

Now the distance of a point on parabola from focus at #(6,5)# is always equal to its distance from the directrix.

Let this point be #(x,y)#.

Its distance from focus is #sqrt((x-6)^2+(y-5)^2)# and from directrix will be #|y-19|#

Hence, #(x-6)^2+(y-5)^2=(y-19)^2#

or #x^2-12x+36+y^2-10y+25=y^2-38y+361#

or #x^2-12x+28y-300=0#

As vertex form of equation is #y=a(x-h)^2+k#

the above is #28y=-x^2+12x+300# or #28y=-(x^2-12x+36-336)#

or #28y=-(x-6)^2+336# or

#y=-1/28(x-6)^2+336/28# or

#y=-1/28(x-6)^2+12#

graph{-1/28(x-6)^2+12 [-12.83, 27.17, -2.08, 17.92]}