How do you write the equation given vertex (8,6) and focus (2,6)?

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Answer 1 · 2017-04-16T06:28:21

#x=-1/24(y-6)^2+8#

As the vertex is #(8,6)# and focus is #(2,6)#, axis of symmetry is #y=6#.

And as directrix is perpendicular to axis of symmetry and vertex is equidistant from focus and directrix, equation of directrix is #x=14#.

Now parabola is locus of a point that moves so that it's distance from focus #(2,6)# and directrix is always same. Hence equation of parabola is

#(x-2)^2+(y-6)^2=(x-14)^2#

or #x^2-4x+4+y^2-12y+36=x^2-28x+196#

or #y^2-12y-156=-24x#

or #-24x=y^2-12y+36-192#

or #x=-1/24(y-6)^2+8#

graph{y^2-12y+24x+156=0 [-125.5, 34.5, -33.5, 46.5]}