1985 AIME Problems/Problem 11
Problem
An ellipse has foci at and
in the
-plane and is tangent to the
-axis. What is the length of its major axis?
Solution
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. Finding the optimal location for
is a classic problem: for any path from
to
and then back to
, we can reflect 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.
![[asy] size(200); pointpen=black;pathpen=black+linewidth(0.6);pen f = fontsize(10); pair F1=(9,20),F2=(49,55); D(shift((F1+F2)/2)*rotate(41.186)*scale(85/2,10*11^.5)*unitcircle); D((-20,0)--(80,0)--(0,0)--(0,80)--(0,-60)); path p = F1--(49,-55); pair X = IP(p,(0,0)--(80,0)); D(p,dashed);D(F1--X--F2);D(F1);D(F2);D((49,-55)); MP("X",X,SW,f); MP("F_1",F1,NW,f); MP("F_2",F2,NW,f); MP("F_2'",(49,-55),NE,f); [/asy]](http://latex.artofproblemsolving.com/e/1/7/e17edad4abef8ffedba4e8086dacea0dc6c8c309.png)
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
.
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |