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

1 Answer
Jun 13, 2016

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