Question

Find the equation of the parabola whose vertex is $(3,-2)$ and focus is $(3,1)$.

Answer 1

Clearly we can see that axis of parabola is parallel to y-axis

Assume the equation of parabola to be :$(x-h{)}^2=4b(y-k)$

Focus =$(h,k+b)=(3,1)$

Vertex $=(h,k)=(3,-2)$

$\Rightarrow h=3,k=-2,b=3$

$\Rightarrow$equation of parabola is : $(x-3{)}^2=12(y+2)$