Question

How do you find the equation in standard form of the parabola Focus (3,4) and a directrix y = 1?

Answer 1

`x^2-6x-6y+24=0`

Explanation:

We use Focus-Directrix Property of Parabola :
If Focus is `S(h,k)` & eqn. of Directrix is `ax+by+c=0`, eqn. of parabola is `(x-h)^2+(y-k)^2=[(ax+by+c)/{sqrt(a^2+b^2)}]^2= (ax+by+c)^2/(a^2+b^2)`

In our case, it is,
`(x-3)^2+(y-4)^2=(0*x+1y-1)^2/(0^2+1^2)=(y-1)^2`
`:. x^2-6x+9-8y+16+2y-1=0`
`:. x^2-6x-6y+24=0`