How to eliminate $\theta$? While doing a sum I was stuck in a particular step:
$$r_1= \frac{4a \cos \theta }{\sin^2 \theta}$$ and $$r_2=\frac{4a \sin \theta }{\cos^2 \theta}$$
How to eliminate $\theta$ ?

$$\begin{cases}
\displaystyle
r_1=\frac{4a\cos\theta}{\sin^2\theta}\\
\displaystyle
r_2=\frac{4a\sin\theta}{\cos^2\theta}\\
\end{cases}$$
Try to find $\sin\theta$ and $\cos\theta$ as following:
$$r_1\sin^2\theta=4a\cos\theta\tag{1}$$
$$r_2\cos^2\theta=4a\sin\theta\tag{2}$$
To equation $(1)$ plug computed $\sin\theta$ from equation $(2)$:
$$\sin\theta=\frac{r_2\cos^2\theta}{4a}$$
$$r_1\left(\frac{r_2\cos^2\theta}{4a}\right)^2=4a\cos\theta$$
$$r_1\frac{r_2^2\cos^4\theta}{16a^2}=4a\cos\theta$$
$$r_1r_2^2\cos^3\theta=64a^3$$
$$\cos^3\theta=\frac{64a^3}{r_1r_2^2}$$
$$\cos\theta=\sqrt[3]{\frac{64a^3}{r_1r_2^2}}=\frac{4a}{\sqrt[3]{r_1r_2^2}}$$
Same thing to compute $\sin\theta$:
$$\sin\theta=\frac{4a}{\sqrt[3]{r_2r_1^2}}$$
And use the Pythagorean identity:
$$\sin^2\theta+\cos^2\theta=1$$
$$\left(\frac{4a}{\sqrt[3]{r_2r_1^2}}\right)^2+\left(\frac{4a}{\sqrt[3]{r_1r_2^2}}\right)^2=1$$
$$\frac{16a^2}{(r_2r_1^2)^{2/3}}+\frac{16a^2}{(r_1r_2^2)^{2/3}}=1$$
