1988 AIME Problems/Problem 14
Problem
Let be the graph of , and denote by the reflection of in the line . Let the equation of be written in the form
Find the product .
Solution 1
Given a point on , we look to find a formula for on . Both points lie on a line that is perpendicular to , so the slope of is . Thus . Also, the midpoint of , , lies on the line . Therefore .
Solving these two equations, we find and . Substituting these points into the equation of , we get , which when expanded becomes .
Thus, .
Solution 2
The asymptotes of are given by and . Now if we represent the line by the complex number , then we find the direction of the reflection of the asymptote by multiplying this by , getting . Therefore, the asymptotes of are given by and .
Now to find the equation of the hyperbola, we multiply the two expressions together to get one side of the equation: . At this point, the right hand side of the equation will be determined by plugging the point , which is unchanged by the reflection, into the expression. But this is not necessary. We see that , , so .
Solution 3
The matrix for a reflection about the polar line is: This is not hard to derive using a basic knowledge of linear transformations. You can refer here for more information: https://en.wikipedia.org/wiki/Orthogonal_matrix
Let . Then , so and . Therefore, if is mapped to under the reflection, then and .
The new coordinates must satisfy . Therefore, Thus, and , so . The answer is .
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.