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 $a=x^2$ 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
$\fbox{C}$