Difference between revisions of "1988 AIME Problems/Problem 14"
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Find the product <math>bc</math>. | Find the product <math>bc</math>. | ||
− | == Solution == | + | == Solution 1 == |
Given a point <math>P (x,y)</math> on <math>C</math>, we look to find a formula for <math>P' (x', y')</math> on <math>C^*</math>. Both points lie on a line that is [[perpendicular]] to <math>y=2x</math>, so the slope of <math>\overline{PP'}</math> is <math>\frac{-1}{2}</math>. Thus <math>\frac{y' - y}{x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y</math>. Also, the midpoint of <math>\overline{PP'}</math>, <math>\left(\frac{x + x'}{2}, \frac{y + y'}{2}\right)</math>, lies on the line <math>y = 2x</math>. Therefore <math>\frac{y + y'}{2} = x + x' \Longrightarrow 2x' - y' = y - 2x</math>. | Given a point <math>P (x,y)</math> on <math>C</math>, we look to find a formula for <math>P' (x', y')</math> on <math>C^*</math>. Both points lie on a line that is [[perpendicular]] to <math>y=2x</math>, so the slope of <math>\overline{PP'}</math> is <math>\frac{-1}{2}</math>. Thus <math>\frac{y' - y}{x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y</math>. Also, the midpoint of <math>\overline{PP'}</math>, <math>\left(\frac{x + x'}{2}, \frac{y + y'}{2}\right)</math>, lies on the line <math>y = 2x</math>. Therefore <math>\frac{y + y'}{2} = x + x' \Longrightarrow 2x' - y' = y - 2x</math>. | ||
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Thus, <math>bc=(-7)(-12)=\boxed{084}</math>. | Thus, <math>bc=(-7)(-12)=\boxed{084}</math>. | ||
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+ | == Solution 2 == | ||
+ | The asymptotes 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>. | ||
== See also == | == See also == |
Revision as of 18:47, 9 June 2013
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 |