# A nasty proof: angle bisectors and ellipses

This came up in class, and took me several attempts, so I thought I’d share it. The question asks about an ellipse with equation $9x^2 + 25y^2 = 225$, with foci $A$ and $B$ at $(\pm4,0)$ - the challenge is to prove that the normal to the ellipse at a point $P$ bisects the angle $APB$.

It’s a vile, vile question, and not something that should be left alone near teenagers.

### Here’s the proof. Please make sure your seatbelt is securely fastened.

The formula book give you: the acute angle between two lines is $\arctan \left| \frac{M-m}{1+Mm} \right|$ (*).

If you change to polar co-ordinates, $P$ is at $(5 \cos(t), 3 \sin(t))$. We’re given that $A$ is at $(-4,0)$ and $B$ is at $(4,0)$; we can work out that $X$, where the normal crosses the axis, is at $\left(\frac{16}{5}\cos(t), 0\right)$. ((Left as an exercise.))

Let’s also define $S = \sin(t)$ and $C = \cos(t)$ to save my keyboard.

We want to show that $APX = XPB$ or that $\tan(APX) = \tan(XPB)$. (This is equivalent as long as both angles are acute, and they are).

The gradient of $AP$ is $m_a = \frac{3S}{5C + 4}$.

The gradient of $XP$ is $m_b = \frac{3S}{5C - \frac{16}{5}C} = \frac{5S}{3C}$.

The gradient of $BP$ is $m_c = \frac{3S}{5C - 4}$.

Using (*), $\tan(APX) = \frac{ \frac{3S}{5C + 4} - \frac{5S}{3C} }{1 + \frac{3S}{5C + 4} \frac{5S}{3C} }$, and $\tan(XPB) = \frac{ \frac{5S}{3C} - \frac{3S}{5C - 4} }{1 + \frac{3S}{5C - 4} \frac{5S}{3C} }$.

If those two messes are equal, we’re done!

Let’s multiply $\tan(APX)$ top and bottom by $3C(5C+4)$ to get $\frac{ {9SC} - {5S(5C+4} }{3C(5C+4) + 15S^2 } = \frac{-16SC - 20S}{15 + 12C} = \frac{-4S(4C+5)}{3(5+4C} = -\frac43 S$.

Similarly for $\tan(XPB)$, we get $\frac{ 5S(5C-4) - 9 CS }{(5C-4)(3C) + 15S^2} = \frac{16SC - 20S}{15 - 12C} = \frac{4S(4C - 5)}{3(5-4C)} = -\frac43 S = \tan(APX)$.

I think that deserves a $\blacksquare$.