1967 AHSME Problems/Problem 14

Revision as of 18:19, 8 October 2015 by Vatatmaja (talk | contribs) (Solution)

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

then

        $a+4b=1$

and

        $4a+b=4$

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


$=>$ $\fbox{C}$

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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