Difference between revisions of "1988 AIME Problems/Problem 14"
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− | The [[asymptote]] of <math>C</math> are given by <math>x=0</math> and <math>y=0</math>. Now if we represent the line <math>y=2x</math> by the complex number <math>1+2i</math>, then we find the direction of the reflection of the asymptote <math>x=0</math> by multiplying this by <math>2-i</math>, getting <math>4+3i</math>. Therefore, the asymptotes of <math>C^*</math> are given by <math>4y-3x=0</math> and <math>3y+4x=0</math>. | + | The [[asymptote]]s of <math>C</math> are given by <math>x=0</math> and <math>y=0</math>. Now if we represent the line <math>y=2x</math> by the complex number <math>1+2i</math>, then we find the direction of the reflection of the asymptote <math>x=0</math> by multiplying this by <math>2-i</math>, getting <math>4+3i</math>. Therefore, the asymptotes of <math>C^*</math> are given by <math>4y-3x=0</math> and <math>3y+4x=0</math>. |
Now to find the equation of the hyperbola, we multiply the two expressions together to get one side of the equation: <math>(3x-4y)(4x+3y)=12x^2-7xy-12y^2</math>. At this point, the right hand side of the equation will be determined by plugging the point <math>(\frac{\sqrt{2}}{2},\sqrt{2})</math>, which is unchanged by the reflection, into the expression. But this is not necessary. We see that <math>b=-7</math>, <math>c=-12</math>, so <math>bc=\boxed{084}</math>. | Now to find the equation of the hyperbola, we multiply the two expressions together to get one side of the equation: <math>(3x-4y)(4x+3y)=12x^2-7xy-12y^2</math>. At this point, the right hand side of the equation will be determined by plugging the point <math>(\frac{\sqrt{2}}{2},\sqrt{2})</math>, which is unchanged by the reflection, into the expression. But this is not necessary. We see that <math>b=-7</math>, <math>c=-12</math>, so <math>bc=\boxed{084}</math>. |
Revision as of 01:22, 16 February 2014
Contents
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
.
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.