Mock Geometry AIME 2011 Problems/Problem 14
The point is a focus of a non-degenerate ellipse tangent to the positive and axes. the locus of the center of the ellipse lies along graph of, where are positive integers with no common factor other than . Find
We utilize (the proof of) a highly useful lemma.
LEMMA: Let be an ellipse with foci and , and let be a point on . Furthermore, let be the line tangent to at . Then the acute angles formed by and with respect to are equal. (This is analogous to the theorem stating that the angle of incidence is equal to the angle of reflection in optics.)
Proof. Let be the length of the semimajor axis of the ellipse. Denote by the reflection of across . Remark that , which is fixed. Furthermore, for any other point on , the distance is intuitively larger than (since a "bigger" ellipse is needed to pass through ). Hence is the point for which is minimized, and so , , are collinear. Thus the two acute vertical angles formed by and are congruent - tracing back the reflection gives our desired. (For more detail, see http://www.maa.org/sites/default/files/0746834207514.di020724.02p0009e.pdf .)
This reflecting business motivates what to do next. Let and be the reflections of across the and axes respectively, and let be the location of the other focus of our ellipse. Note that said axes are tangent to the ellipse, so we are basically replicating the proof above. Now recall that by the proof and both equal , so they are equal in length to each other. Thus the locus of points is the perpendicular bisector of segment . To find the equation of this perpendicular bisector, remark that the midpoint of is the origin of the plane and that the slope of is , so the equation is .
Finally, we compute the locus of the centers. We could go through a slight coordinate bash (and it isn't that hard to do), but here we shall use Euclidean geometry. Consider the homothety centered at with scale factor . For an arbitrary point on this line, said homothety will take to the midpoint of line segment - which is just the center of the ellipse with foci and ! Hence the desired locus of centers is the line that results when the entire line is transformed under . By basic homothety rules, the line is parallel to . Furthermore, this line must pass through the midpoint of the segment connecting the origin to , or the point . Thus, the equation of the desired locus is Our requested answer is .