Difference between revisions of "1967 AHSME Problems/Problem 14"

Revision as of 17:19, 8 October 2015

The number of distinct points common to the curves $x^2+4y^2=1$ and $4x^2+y^2=4$ is:

$\text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 3 \quad \text{(E) } 4$

Let $x^2 = a$ and $y^2 = b.$

then

and

By solving we find--

$a=1$

$b=0$

However $a=x^2$ and $b=y^2$

Therefore $y=0$ , and $x=1,-1$

Thus, the only solutions are $(0,1)$ , and $(0,-1)$

So there are only 2 solutions

$=>$ $\fbox{C}$

1967 AHSME (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

@@ Line 5: / Line 5: @@
 == Solution ==
-<math>\fbox{C}</math>
+Let <math>x^2 = a</math> and <math>y^2 = b.</math>
+then
+         <math>a+4b=1</math>
+and
+         <math>4a+b=4</math>
+By solving we find--
+<math>a=1</math>
+<math>b=0</math>
+However <cmath>a=x^2</cmath>
+and <cmath>b=y^2</cmath>
+Therefore
+<math>y=0</math>, and <math>x=1,-1</math>
+Thus, the only solutions are <math>(0,1)</math>, and <math>(0,-1)</math>
+So there are only 2 solutions
+<math>=></math> <math>\fbox{C}</math>
 == See also ==