Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 3"
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Through expanding the second equation, we get <math>-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 1</math>. Since <math>2y^2-x^2 = 1</math>, we have <cmath>-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 2y^2-x^2</cmath> | Through expanding the second equation, we get <math>-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 1</math>. Since <math>2y^2-x^2 = 1</math>, we have <cmath>-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 2y^2-x^2</cmath> | ||
− | The <math>x^2</math> terms on each side cancel out, so the equation becomes < | + | The <math>-x^2</math> terms on each side cancel out, so the equation becomes |
+ | <cmath>(8B-24)xy + (2B^2-16)y^2 = 2y^2</cmath> | ||
+ | The coefficient of <math>xy</math> on the RHS is <math>0</math> and the coefficient of <math>y^2</math> is <math>2</math>. From these two observations, we now create two new equations. | ||
<cmath>8B-24 = 0</cmath> | <cmath>8B-24 = 0</cmath> | ||
<cmath>2B^2-16 = 2</cmath> | <cmath>2B^2-16 = 2</cmath> |
Latest revision as of 03:18, 11 June 2022
Problem
Find all values of that have the property that if
lies on the hyperbola
,
then so does the point
.
Solution 1
We can write a system of equations -
Expanding the second equation, we get .
Since we want this to look like , we plug in B's that would put it into that form. If we plug in
, things cancel, and we get
. So
~Ultraman
Solution 2 (Grinding)
As with Solution 1, we create a system of equations.
Through expanding the second equation, we get . Since
, we have
The
terms on each side cancel out, so the equation becomes
The coefficient of
on the RHS is
and the coefficient of
is
. From these two observations, we now create two new equations.
Solving either equation and then checking with the other will reveal that
.
~kingme271
See also
2018 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |