Four Leaf Clover

The normal to the ellipse at the point $(a\cos\theta,b\sin\theta)$ has equation $ax\sin\theta - by\cos\theta \; = \; (a^2-b^2)\sin\theta\cos\theta$ which meets the ellipse again at the point $(a\cos\phi,b\sin\phi)$ , where $\begin{aligned} a^2\sin\theta\cos\phi - b^2\cos\theta\sin\phi & = \; (a^2-b^2)\sin\theta\cos\theta \\ \sin\big(\tfrac{\phi-\theta}{2}\big)\left[a^2\sin\theta\sin\big(\tfrac{\phi+\theta}{2}\big) + b^2\cos\theta\cos\big(\tfrac{\phi+\theta}{2}\big)\right] & = \; 0 \\ a^2\tan\theta\tan\big(\tfrac{\phi+\theta}{2}\big) + b^2 & = \; 0 \end{aligned}$ This last equation can be solved to obtain $\tan\tfrac12\phi$ in terms of $\theta$ , from which it is easy to express $\sin\phi$ and $\cos\phi$ in terms of $\theta$ . The midpoint of this chord has coordinates $(X,Y)$ , where $X \; = \; \tfrac12a(\cos\theta+\cos\phi) \; = \; \frac{a^3(a^2-b^2)\sin^2\theta\cos\theta}{a^4\sin^2\theta+b^4\cos^2\theta} \hspace{2cm} Y \; = \; \tfrac12b(\sin\theta+\sin\phi) \; =\; -\frac{b^3(a^2-b^2)\sin\theta\cos^2\theta}{a^4\sin^2\theta+b^4\cos^2\theta}$ Thus the area of the locus is $\tfrac12\int_0^{2\pi}\big[X(\theta)Y'(\theta)-X'(\theta)Y(\theta)\big]\,d\theta \; = \; \int_0^{2\pi} \frac{2a^3b^3(a^2-b^2)^2\sin^2\theta\cos^2\theta}{(a^4+b^4 - (a^4-b^4)\cos4\theta)^2}\,d\theta \; = \; \frac{ab(a^2-b^2)^2\pi}{2(a^2+b^2)^2} \; = \; \frac{a^5be^4\pi}{2(a^2+b^2)^2}$ after some elementary contour integration.