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

Revision as of 14:53, 10 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

$\rightarrow$ $\fbox{C}$

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

Revision as of 18:19, 8 October 2015 (view source) Vatatmaja (talk \| contribs) (→‎Solution) ← Older edit		Revision as of 14:53, 10 October 2015 (view source) Carrots101 (talk \| contribs) m (→‎Solution) Newer edit →
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−	<math>=></math> <math>\fbox{C}</math>	+	<math>\rightarrow</math> <math>\fbox{C}</math>

	== See also ==		== See also ==