Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 3"

(Problem)
m (Solution)
(4 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Find all values of B that have the property that if (x, y) lies on the hyperbola 2y^2-x^2 = 1,
+
Find all values of <math>B</math> that have the property that if <math>(x, y)</math> lies on the hyperbola <math>2y^2-x^2 = 1</math>,
then so does the point (3x + 4y, 2x + By).
+
then so does the point <math>(3x + 4y, 2x + By)</math>.
  
 
== Solution ==
 
== Solution ==
 +
We can write a system of equations -
 +
<cmath>2y^2-x^2 = 1</cmath>
 +
<cmath>2(2x + By)^2 - (3x+4y)^2 = 1</cmath>
  
 +
Expanding the second equation, we get <math>-x^2+8Bxy-24xy+2B^2y^2-16y^2=1</math>.
 +
 +
Since we want this to look like <math>2y^2-x^2=1</math>, we plug in B's that would put it into that form. If we plug in <math>B=3</math>, things cancel, and we get <math>-x^2+24xy-24xy+18y^2-16y^2=1 \rightarrow 2y^2-x^2=1</math>. So <math>\boxed{B=3}</math>
 +
~Ultraman
  
 
== See also ==
 
== See also ==
 
{{UNCO Math Contest box|year=2018|n=II|num-b=2|num-a=4}}
 
{{UNCO Math Contest box|year=2018|n=II|num-b=2|num-a=4}}
  
[[Category:]]
+
[[Category:Introductory Algebra Problems]]

Revision as of 14:53, 2 January 2020

Problem

Find all values of $B$ that have the property that if $(x, y)$ lies on the hyperbola $2y^2-x^2 = 1$, then so does the point $(3x + 4y, 2x + By)$.

Solution

We can write a system of equations - \[2y^2-x^2 = 1\] \[2(2x + By)^2 - (3x+4y)^2 = 1\]

Expanding the second equation, we get $-x^2+8Bxy-24xy+2B^2y^2-16y^2=1$.

Since we want this to look like $2y^2-x^2=1$, we plug in B's that would put it into that form. If we plug in $B=3$, things cancel, and we get $-x^2+24xy-24xy+18y^2-16y^2=1 \rightarrow 2y^2-x^2=1$. So $\boxed{B=3}$ ~Ultraman

See also

2018 UNCO Math Contest II (ProblemsAnswer KeyResources)
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