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

Question

4386 views around the world

You can reuse this answer
Creative Commons License

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]}