# How to find the farthest point on a ellipse from a point within an ellipse?

I was wondering if you could help me figure this out.

I've been trying to write some code to calculate the farthest point on an ellipse $$(150w, 85h)$$ from a given point $$(55x, 20y)$$ within the ellipse. Could anyone help walk me through the steps to achieve this? I've looked through a few examples and it still isn't clicking yet. Is there a way to find the farthest point on the ellipse from a point within the ellipse without using brute force (comparing each point on the ellipse to the point within the ellipse)?

Assume we only know the ellipse size/location, and the point's location.

Also, the numbers I am using are made up so feel free to change them to illustrate your point.

• Ellipse: size $$(150w, 85h)$$, center $$(0x, 0y)$$ (is known)
• Point within ellipse: $$(55x, 20y)$$ (is known)

Thanks!

Edited for clarification. I appreciate the responses so far, but I'm looking for a formula where I don't have to compare the point within the ellipse to each point on the ellipse's edge (ideally).

• The problem should be equivalent to finding the circumference with center in your point and externally tangent to the ellipse. – dfnu Mar 20 at 21:27
• At least $2$ and at most $4$ normals pass through a given point. The furthest one is what you want. See another answer of mine here. – Ng Chung Tak Mar 21 at 1:20
• There’s no simple formula for this. The answer that you’ve gotten explains how to find such a formula, but it will involve solving a trigonometric equation or something else equally unpleasant. – amd Mar 21 at 18:06

HINT

Given a point $$P$$ on the ellipse and a point $$Q$$ not on the ellipse, the distance from $$P$$ to $$Q$$ is a local extremum if the chord $$PQ$$ is perpendicular to the tangent line to the ellipse at $$P$$.

We can parametrise the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ by $$P(a\cos t,b\sin t)$$. Take a point $$Q(u,v)$$ not on the ellipse, then $$\vec{PQ}=(u-a\cos t){\bf i} + (v-b\sin t){\bf j}$$

The tangent to the ellipse at $$P$$ is parallel to $$(-a\sin t){\bf i}+(b\cos t){\bf j}$$.

Using the dot/scalar product, these two are perpendicular if, and only if, $$(-a\sin t)(u-a\cos t) + (b\cos t)(v-b\sin t)=0$$

For fixed $$a$$, $$b$$, $$u$$ and $$v$$, solve this for $$t$$.

• There will generally be multiple solutions to that equation. Is there a way other than brute-force checking to pick out the one that gives the minimum distance? – amd Mar 20 at 21:55
• You'd only expect four solutions for $0 \le t < 2\pi$. You can eliminate two immediately by looking at which quadrant your point is in. That leaves two to check. – Fly by Night Mar 21 at 16:59