## Precalculus (6th Edition) Blitzer

$x=\dfrac{1}{12}(y-4)^2-1$
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, $x=\dfrac{1}{12}(y-4)^2-1$