1988 AIME Problems/Problem 14
Find the product .
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 .
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 .
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 . Note that the line of reflection, , is the polar line . Then , so and .
Therefore, if is mapped to under the reflection, then and . Since the transformation matrix represents a reflection, it must be its own inverse; therefore, and .
The original coordinates must satisfy . Therefore, Thus, and , so . The answer is .
Solution 4(IMO the BEST solution)
Find some simple points on the graph of C, reflect them and note down the new coordinates, and plug 'em into the given equation. After some plugging and chugging and solving the system of equations that follow, we get that and , so . The answer is .
|1988 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|