Question

What is the equation of the parabola with a focus at (3,-8) and a directrix of y= -5?

Answer 1

The equation is `y=-1/6(x-3)^2-39/6`

Explanation:

Any point `(x,y)` on the parabola is equidistant from the directrix and from the focus.

Therefore,

`(y+5)=sqrt((x-3)^2+(y+8)^2)`

Squaring both sides

`(y+5)^2=(x-3)^2+(y+8)^2`

`y^2+10y+25=(x-3)^2+y^2+16y+64`

`6y=-(x-3)^2-39`

`y=-1/6(x-3)^2-39/6`

graph{(y+1/6(x-3)^2+39/6)(y+5)=0 [-28.86, 28.87, -14.43, 14.45]}

Answer 2

The equation of parabola is `y=-1/6(x-3)^2-6.5`

Explanation:

Focus is at `(3,-8)`and directrix is `y=-5`. Vertex is at midway

between focus and directrix. Therefore,vertex is at `(3,(-5-8)/2)`

or at `(3, -6.5)` . The vertex form of equation of parabola is

`y=a(x-h)^2+k ; (h,k)` being vertex. `h=3 and k = -6.5`

So the equation of parabola is `y=a(x-3)^2-6.5`. Distance of

vertex from directrix is `d= |6.5-5|=1.5`, we know `d = 1/(4|a|)`

`:. 1.5 = 1/(4|a|) or |a|= 1/(1.5*4)=1/6`. Here the directrix is above

the vertex , so parabola opens downward and `a` is negative.

`:. a= -1/6`. Hence the equation of parabola is

`y=-1/6(x-3)^2-6.5`

graph{-1/6(x-3)^2-6.5 [-40, 40, -20, 20]}