1966 AHSME Problems/Problem 17

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Problem

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

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

Solution

Let $a=x^2$ and $b=y^2$. We now have a system of 2 linear equations: $a+4b=1\\4a+b=4$ Multiplying the first equation by 4 and then subtracting the second equation from the first one, we get: $15b=0\\b=0$ Now, we can substitute b to solve for a: $4a+0=4\\a=1$ Now note that $x^2=1 \rightarrow x=\pm 1$ and $y^2=0\rightarrow y=0$, so the solutions are $(1,0), (-1, 0)$. There are two solutions, meaning that the answer is $\fbox{C}$

See also

 1966 AHSME (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 17 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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