Question

How do you write the equation of the parabola in vertex form given vertex (5,-2) and focus (5,-4)?

Answer 1

Equation is `y=-1/8(x-5)^2-2`

Explanation:

As vertex is `(5,-2)` and focus is `(5,-4)`

as axis of symmetry passes through both and abscissa is common in both, equation of axis of symmetry is `x=5`

and as directrix is perpendicular to it, its equation is of type `y=k`

Now vertex is midway between directrix and focus and hence directrix is `y=0`

The parabola is locus of a point `(x,y)` which moves so that it is equidistant from directrix anf focus, hence its equation is

`(x-5)^2+(y+4)^2=y^2`

or `(x-5)^2+y^2+8y+16=y^2`

or `8y=-(x-5)^2-16`

i.e. `y=-1/8(x-5)^2-2`

graph{y(x^2-10x+8y+41)(x-5)((x-5)^2+(y+2)^2-0.03)((x-5)^2+(y+4)^2-0.03)=0 [-5.04, 14.96, -7.16, 2.84]}