Answer
$x=\dfrac{1}{12}(y-4)^2-1$
Work Step by Step
General form of a horizontal parabola is given as: $(y-k)^2=4p(x-h)$...(1)
Here, $\text{Vertex}=(h,k)$ and focus is: $(h+p, k)$
General form of a vertical parabola is given as: $(x-h)^2=4p(y-k)$...(2)
Here, $\text{Vertex}=(h,k)$ and focus is: $(h, k+p)$
As we are given focus: $(2,4)$
Thus, from both forms of a parabola , we have two equations$h-p=-4$ and $h+p=2$
Add these two equations, we get: $h=-1$
Thus, $p=2-h=2-(-1)=3$
From equation (1), we have $(y-4)^2=4(3)(x-(-1))$
or, $(y-4)^2=12(x+1)$
or, $x+1=\dfrac{(y-4)^2}{12}$
or, $x=\dfrac{1}{12}(y-4)^2-1$
