Rectangle inscribed on two parabolas but not parallel to the axis Let $f(x)=2x^2$ and $g(x)=36-x^2$. Let $R$ be the closed region between them. Draw a rectangle $ABCD$ such that $A$ and $B$ are points on $f(x)$, and $C$ and $D$ are points on $g(x)$, and the rectangle is not parallel to the axis, or prove that to be impossible. 

Brute force:
First, we relax the condition to finding any parallelogram with 2 vertices on each equation.
Let $ A = (a,2a^2), B = (a+b, 2a^2 + 4ab + 2b^2), D = (d, 36-d^2)$, which makes $ C = (d+b, 36-(d+b)^2 = 36 - d^2 + 4ab + 2b^2)$. This gives $ b=0$(rejected), $d =-2a- \frac{3b}{2}$.
Now, we to make the parallelogram into a rectangle, we require $ AB \perp AD \Rightarrow (b, 4ab + 2b^2) \cdot (d-a, 36 - d^2 - 2a^2) = 0$. This simplifies to   
$$-3/2 b (2 a + b) (8 a^2 + 8 a b + 3 b^2 - 47) = 0$$
Note that $b=0$ is rejected, and $2a+b$ means $AB$ is parallel to the x-axis, hence rejected. So, we're looking for solutions to   $8 a^2 + 8 a b + 3 b^2 - 47=0$, which is an ellipse.   
As it turns out, $ a = -3, b = 4 - \sqrt{\frac{23}{3}}$ is one such solution, which leads to the rectangle.  
$$ A = ( -3, 18), B \approx ( -1.77, 6.26), C\approx (5.38, 7.01),D \approx (4.15, 18.75)$$
Of course, if you don't want $A$ on the intersection point, take $a = -3 + \epsilon$. 
