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I have an ellipse centered at $(h,k)$, with semi-major axis $r_x$, semi-minor axis $r_y$, both aligned with the Cartesian plane.

How do I determine if a point $(x,y)$ is within the area bounded by the ellipse?

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Which of the two solutions is more efficient (computationally-wise) assuming that in both cases the "|x−h|>rx" and "|y−k|>ry" rejections are implemented? –  Ricardo Sánchez-Sáez Oct 15 '12 at 13:19
@rsanchezsaez Probably Srivatsan's because square roots are slow; if you're concerned about performance writing both and benching them is probably the best route. Alternately, you could try posing the question on Stack Overflow. –  Dan Neely Oct 15 '12 at 14:27
Thank for the tips Dan. I implemented Srivatsan's and seems to work fine. I don't want to spend much time in premature optimization. If in the future we run into performance issues we will profile to see if this is the bottleneck. –  Ricardo Sánchez-Sáez Oct 15 '12 at 15:00
"Aligned with the Cartesian plane" means that the major and minor axes lie on the coordinate axes, right? –  rschwieb Feb 5 '13 at 17:41
@rschwieb yes, it does. –  Dan Neely Feb 5 '13 at 18:11
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3 Answers

up vote 17 down vote accepted

The region (disk) bounded by the ellipse is given by the equation: $$ \frac{(x-h)^2}{r_x^2} + \frac{(y-k)^2}{r_y^2} \leq 1. \tag{1} $$ So given a test point $(x,y)$, plug it in $(1)$. If the inequality is satisfied, then it is inside the ellipse; otherwise it is outside the ellipse. Moreover, the point is on the boundary of the region (i.e., on the ellipse) if and only if the inequality is satisfied tightly (i.e., the left hand side evaluates to $1$).

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I have just registered to upvote this. I am developing app for android and this has been EXACTLY what I was looking for. Thank u –  Eugene Apr 4 '13 at 7:56
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Another way uses the definition of the ellipse as the points whose sum of distances to the foci is constant.

Get the foci at $(h+f, k)$ and $(h-f, k)$, where $f = \sqrt{r_x^2 - r_y^2}$.

The sum of the distances (by looking at the lines from $(h, k+r_y)$ to the foci) is $2\sqrt{f^2 + r_y^2} = 2 r_x $.

So, for any point $(x, y)$, compute $\sqrt{(x-(h+f))^2 + (y-k)^2} + \sqrt{(x-(h-f))^2 + (y-k)^2} $ and compare this with $2 r_x$.

This takes more work, but I like using the geometric definition.

Also, for both methods, if speed is important (i.e., you are doing this for many points), you can immediately reject any point $(x, y)$ for which $|x-h| > r_x$ or $|y-k| > r_y$.

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Right, it's easier to check if a point is within a rectangle. If it isn't in the rectangle, it certainly can't be in the ellipse. –  J. M. Oct 28 '11 at 7:16
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I have another solution. In summary, transform everything so you can test whether a point is within a circle centered at (0,0). Since the ellipse is oriented orthogonally to the Cartesian plane, we can simply scale one of the dimensions by the quotient of the two axes.

First subtract (h,k) from both points.

(h,k) becomes (0,0). (x,y) becomes (x-h,y-k).

Now we scale the second coordinate, normalizing it to the first.

Scaling (x-h,y-k) we get $$ (x-h,(y-k) * \frac{r_x}{r_y}) $$

Now we simply need to test if $$ |(x-h,(y-k) * \frac{r_x}{r_y})| <= r_x $$

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I apologize. The coding of the fractions didn't take. Perhaps someone can help me out with that and point me to a reference? I was using meta.math.stackexchange.com/questions/5020/… as a guide. Edit: I figured it out. I was missing the \$$ tag. –  Victor Engel Nov 24 '12 at 3:45
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