Area of Parallelogram in an Ellipse

Question

A parallelogram is inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with the fixed line $y=mx$ as one of its diagonals. Prove that the maximum area of the parallelogram is $2ab$.

First I found out the the points of intersection of the line with the ellipse by putting $y=mx$ in the ellipse equation. This is what I got, $$\frac{x^2}{a^2}+\frac{m^2x^2}{b^2}=1$$ $$\implies x^2(\frac{1}{a^2}+\frac{m^2}{b^2})=1$$ $$\implies x=\pm\frac{ab}{\sqrt{a^2m^2+b^2}}$$

This gives me, $$ y=\pm\frac{mab}{\sqrt{a^2m^2+b^2}}$$

So, I got the two points where the line intersects the ellipse. I don't know hoew to proceed from here. Do I have to find the point on the ellipse which is at a maximum distance from the line? If yes, how? If no, then how do I solve it?

Answer 1

Do a little diagram of a canonical ellipse, and get convinced that, if the intersection points you found are $\;(\pm x_0,\pm y_0)\;,\;\;x_0,y_0>0$ , then the parallelogram's area is just

$$S(m)=(2x_0)(2y_0)=\frac{4a^2b^2m}{a^2m^2+b^2}$$

Differentiate wrt $\;m\;$ :

$$S'(m)=\frac{4a^2b^2(b^2-a^2m^2)}{(a^2m^2+b^2)^2}=0\iff m=\pm\frac ba$$

It's not hard to see we get a maximum point at $\;m=\frac ba\;$, and thus the area is

$$S\left(\frac ba\right)=\frac{4a^2b^2\frac ba}{a^2\frac{b^2}{a^2}+b^2}=\frac{4ab^3}{2b^2}=2ab$$

Added: The above assumes the parallelogram is a rectangle with sides parallel to the axis. Read the comments below this answer.

Answer 2

Draw the diagonal in the ellipse, and apply an afine transformation changing the ellipse to a circle. Of all the rectangles in this circle sharing the transformed diagonal, that with the largest area will be a square, with area $2r^2$, where $r$ is the radius of the transformed original ellipse. Now afine transform back, and you have a parallelogram with the specified diagonal, and area $2ab$. Note that afine transformations preserve relative areas. Edit: since the area of an inscribed square is larger than any non square quadrilateral inscribed in the circle, the original problem can consider all inscribed quadrilaterals sharing the given diagonal, not jut parallelograms. Further, the two diagonals of the optimum parallelogram are conjugate diameters.