A line through the focus of a parabola intersects the parabola at two points $P$ and $Q$. Show that the tangent line through $P$ is perpendicular...

Question

I need some help with this exercise

"A line through the focus of a parabola intersects the parabola at two points $P$ and $Q$. Show that the tangent line through $P$ is perpendicular to the tangent line through $Q$."

This is the drawing for illustrating this scenario.

The proof is:

$A=A_1$ and $B=B_1$ by the reflection property of parabola

$A=A_2$ by corresponding angles

$B=B_2$ by vertical angles

$A_1=A_2$ and $B_1=B_2$ by the transtive property

By the triangle angles sum theorem:

$A_2+B_2+C=180^{\circ}$

and

$A_1+B_1+D=180^{\circ}$

By sustitution,

$A_1+B_1+C=180^{\circ}$

and

$A_1+B_1+D=180^{\circ}$

Multiplying one of them by $-1$,

$A_1+B_1+C=180^{\circ}$

and

$-A_1-B_1-D=-180^{\circ}$

By elimination,

$C-D=0$

$C=D$

I'm good till here.

The proof finish:

$C=D=90^{\circ}$

How do we get that conclusion?

Answer 1

Nevermind. I just realized that angle $C$ and $D$ form a linear pair and because of that:

$C+D=180^{\circ}$

By sustitution,

$C+C=180^{\circ}$

$2C=180^{\circ}$

$\frac{2C}{2}=\frac{180^{\circ}}{2}$

$C=90^{\circ}$

Therefore, $C=D=90^{\circ}$ and tangent line through $P$ is perpendicular to the tangent line through $Q$.

If there is something wrong in my analysis please let me know. Thanks.

Answer 2

bdvg2302, I think your proof is flawless.

However it can be shortened a bit as follows:

In the figure, notice that the dotted lines are parallel.

This implies that $$2 \alpha + 2 \beta = 180^{\text o}$$ $$\alpha + \beta = 90^{\text o}$$ $$\therefore \gamma = \alpha + \beta = 90^{\text o}$$