How to show that a tangent line to a parabola is a perpendicular bisector

Question

Let $P$ be the parabola given by the equation $P(x)=\frac{1}{4}x^2$ and $a\in\mathbf{R}$.

Let $F=(0,1)$ be the focus of $P$ and $R=(a,P(a))$ be a point on $P$ and let be the point $V=(a,-1)$.

I want to prove that the line $\ell$ tangent to $R$ is the perpendicular bisector of the segment FV by using calculus.

I was able to prove that the line $k$ through $F$ and $V$ has equation $k\colon y=1-\frac{2x}{a}$ and line $\ell$ has equation $\ell\colon y=\frac{1}{4}a(2x-a)$.

How can I complete the proof? I was able to find that $k$ and $\ell$ intersect at the point $S=(\frac{1}{2}a,0)$.

Answer 1

Two parts: perpendicular and bisector.

To show that it is perpendicular, show that the product of the slopes is -1.

To show that it is a bisector, show that the point of intersection is the midpoint of the segment.