Proof involving showing the angles created by a parabola, a tangent, a line parallel to the axis of parabola, and the parabola's focus are equal.

Question

I have to prove that angle $ \alpha $ is equal to angle $ \beta $. The book suggested to show that an isosceles triangle exists. I was able to show that $FP$ = y + p ,where p is the directed distance from the focus and y is the y ordinate for the point of intersection, and $ FQ $ = p - b , where b is the y-intercept for the tangent line. However, I'm not able to make $PQ$ equal to either one of these lengths. Also I don't understand how creating a isosceles will lead to showing that $ \alpha $ = $ \beta $

I added photos.

Answer 1

Note the the focus $F$ has cooridnates $(0,\frac{1}{4a})$.

Slope of the tangent at point $P(x, ax^2)$ is $y'=2ax$ so the equation of the tangent is:

$$Y-y_P=(X-x_P)y'$$

$$Y-ax^2=(X-x)2ax$$

This tangent intersects vertical axis at point $Q(X=0, Y)$ so obviously $Y=-ax^2$. Therefore point $Q$ has the following coordinates: $Q(0, -ax^2)$.

Now:

$$QF=|QO|+OF=ax^2+\frac1{4a}$$

On the other side:

$$PF=\sqrt{(x_P-x_F)^2+(y_P-y_F)^2}=\sqrt{x^2+(ax^2-\frac{1}{4a})^2}$$

$$PF=\sqrt{x^2+a^2x^4-\frac{x^2}2+\frac{1}{16a^2}}= \sqrt{a^2x^4+\frac{x^2}2+\frac{1}{16a^2}}$$

$$PF=\sqrt{(ax^2+\frac1{4a})^2}=ax^2+\frac1{4a}$$

So obviously $QF=PF$ and triangle QPF is isosceles. It follows immediatelly that:

$$\angle QPF=\alpha=\angle PQF=\beta$$