# Equation of a parabola with vertex $V$ and point $P$

Find the equation of the parabola which has the given vertex $$V$$, which passes through the given point $$P$$, and which has the specified axis of symmetry.

$$V(4,-2), P(2,14)$$, vertical axis of symmetry.

The answer is $$(x-4)^2=\frac 14(y+2)$$, but I do not know how to get this answer. I know it is an upward facing parabola with $$x=4$$ as the axis of symmetry.

You are trying to find a parabola of the form $$(x-4)^2 = 4p (y+2)$$ If you replace $x=2$ and $y=14$ en the last equation, you will find $p=\frac{1}{16}$. So that, $$(x-4)^2 = \frac{1}{4} (y+2)$$ is the parabola you are looking for.