1985 AIME Problems/Problem 11
An ellipse is defined to be the locus of points such that the sum of the distances between and the two foci is constant. Let , and be the point of tangency of the ellipse with the -axis. Then must be the point on the axis such that the sum is minimal. (The last claim begs justification: Let be the reflection of across the -axis. Let be where the line through and intersects the ellipse. We will show that . Note that since is on the -axis. Also, since the entire ellipse is on or above the -axis and the line through and is perpendicular to the -axis, we must have with equality if and only if is on the -axis. Now, we have But the right most sum is the straight-line distance from to and the left is the distance of some path from to ., so this is only possible if we have equality and thus ). Finding the optimal location for is a classic problem: for any path from to and then back to , we can reflect (as above) the second leg of this path (from to ) across the -axis. Then our path connects to the reflection of via some point on the -axis, and this path will have shortest length exactly when our original path has shortest length. This occurs exactly when we have a straight-line path, and by the above argument, this path passes through the point of tangency of the ellipse with the axis.
The sum of the two distances and is therefore equal to the length of the segment , which by the distance formula is just .
Finally, let and be the two endpoints of the major axis of the ellipse. Then by symmetry so (because is on the ellipse), so the answer is .
Solution 2 (Calculus)
An ellipse is defined as the set of points where the sum of the distances from the foci to the point is fixed. The length of major axis is equal to the sum of these distances . Thus if we find the sum of the distances, we get the answer. Let k be this fixed sum; then we get, by the distance formula:
This is the equation of the ellipse expressed in terms of . The line tangent to the ellipse at the given point will thus have slope . Taking the derivative gives us the slope of this line. To simplify, let and . Then we get:
Next, we multiply by the conjugate to remove square roots. We next move the resulting form expression into form .
We know and . Simplifying yields:
To further simplify, let and . This means . Solving yields that . Substituting back and yields:
Solving for yields . Substituting back into our original distance formula, solving for yields .
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