Normals to an ellipse: a Core 4 masterclass

A reader (not, in fact, a Core 4 student) wrote in to ask:

I have an ellipse in my spreadsheet program, using the formula $y = \frac ba \sqrt{a^2 - x^2}$, and I want to know the angle the normal to the ellipse makes with the horizontal at any value of $x$.

Ah, the ellipse, my nemesis. We meet again. Luckily, this time, I have the beating of you.

My preferred way of writing an ellipse equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, because it looks neater, so that's what we're going to roll with in this post.

To find the gradient of a normal, we need to differentiate -- implicitly, the way I've set it up. That gives $\frac{2x}{a^2} + \frac{2y}{b^2} \diff{y}{x} = 0$, or $\diff y x = - \frac {x b^2}{y a^2}$ as the gradient of the tangent, so the normal gradient is $m = \frac{y a^2}{x b^2}$.

(You could also do it explicitly with the chain rule: $\diff y x = \frac ba (2x) \left( -\frac 12 (a^2 - x^2)^{-\frac 12}\right) = \frac {bx}{a\sqrt{b^2-x^2}} $. This is equivalent to, although uglier, than the other version.)

So, given the gradient at a point, how do you work out the angle? Simple! To find an angle, you can make a triangle1 and work out -- for instance -- the opposite side divided by the adjacent side, and take the arctangent. In this case, though, opposite divided by adjacent is rise over run -- which is to say, the arctangent of the gradient is your angle!

Here's a Desmos graph to play with!


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. the best shape []


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