# 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 $\left(\frac{16}{5}\cos(t), 0\right)$.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 $\frac{ {9SC} - {5S(5C+4} }{3C(5C+4) + 15S^2 } = \frac{-16SC - 20S}{15 + 12C} = \frac{-4S(4C+5)}{3(5+4C} = -\frac43 S$.

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 $\blacksquare$. ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

1. Left as an exercise. []

### 7 comments on “A nasty proof: angle bisectors and ellipses”

• ##### Joshua Zucker

Is there a better way to show that these two different definitions of an ellipse (a squashed circle, and the reflection property of the foci) are equivalent? Or an easy way to show they’re both equivalent to the sum of the distances version of the definition? That last one seems to have some hope because of some kind of least-time principle, maybe?

• ##### icecolbeveridge

I genuinely don’t know! This is something my student and I battled with for ten days before getting to this… I’d love to see a more elegant proof, if you can find one 🙂

• ##### Fahim Abdullah

Hm but the book says both the tans of both the angles (APX and XPB) should be +4S/3, not -4S/3. And it makes sense too as they’re both acute. :S Although can’t find any mistake here.

• ##### Colin

Good spot – I think I’ve missed out the modulus sign. I wish I could say it was for simplicity, but it was really just carelessness.

• ##### David Richard Getling

Since the angle rotation from the normal to each of the line segments is in opposite directions, getting values of opposite signs seems reasonable.

• ##### David Richard Getling

Thanks for that. I feel deeply ashamed if I come across any A-level question I can’t tackle. My initial approach was using the dot product to get the cosine of the angles, which proved hopeless. I’m not too sure how long it would have taken me to come up with using the tan formula: maybe I wouldn’t.

Now I can get a peaceful night’s sleep.

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