Question

What is the standard form of the equation of the parabola with a focus at (6,5) and a directrix of `y= -1`?

Answer 1

Parabola is `y=1/12(x-6)^2+2`

Explanation:

Parabola is the locus of a point which moves so that its distance from a given point called focus and a given line called directrix is always equal.

Let the point be `(x,y)`. Its distance from focus `(6,5)` is

`sqrt((x-6)^2+(y-5)^2)`

and its distance from directrix `y=-1` is `y+1`

Hence equation of parabola is `sqrt((x-6)^2+(y-5)^2)=y+1`

and squaring `(x-6)^2+(y-5)^2=(y+1)^2`

i.e. `x^2-12x+36+y^2-10y+25=y^2+2y+1`

i.e. `x^2-12x+60=12y`

or `12y=(x-6)^2+24`

or `y=1/12(x-6)^2+2`

graph{(x^2-12x+60-12y)((x-6)^2+(y-5)^2-0.03)(y+1)=0 [-4.08, 15.92, -1.32, 8.68]}