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 $(\pm 4,0)$ - 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 $(\frac{16}{5}\cos (t),0)$. ¹

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 $\frac{9SC-5S(5C+4}{3C(5C+4)+15S^2}=\frac{-16SC-20S}{15+12C}=\frac{-4S(4C+5)}{3(5+4C}=-\frac{4}{3}S$.

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

I think that deserves a $◼$.

Footnotes:

1. Left as an exercise.