The tangents at the extremities of a pair of conjugate diameters form a parallelogram whose area is constant and equal to the product of the axes. The area of parallelogram is

Let $PCQ$ and $RCS$ be two conjugate diameters of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Then the co-ordinates of $P$, $Q$, $R$ and $S$ are

 $P(a\cos \phi ,b\sin \phi )$, $Q(-a\cos \phi ,-b\sin \phi )$,

 $R(-a\sin \phi , b\cos \phi )$ and $S(a\sin \phi ,-b\cos \phi )$ respectively.

The equations of tangents at $P$, $R$, $Q$  and $S$ are

 $\frac{x}{a}\cos \phi +\frac{y}{b}\sin \phi =1$,

$-\frac{x}{a}\sin \phi +\frac{y}{b}\cos \phi =1$, 

$-\frac{x}{a}\cos \phi -\frac{y}{b}\sin \phi =1$,

and $\frac{x}{a}\sin \phi -\frac{y}{b}\cos \phi =1$

Thus, the tangents at $P$ and $Q$ are parallel. Also the tangents at $R$ and $S$ are parallel. Hence, the tangents at $P$, $R$, $Q$, $S$ form a parallelogram.

Area of the parallelogram $=MNM\text{'}N\text{'}$

$=4$ (the area of the parallelogram $CPMR)$

$\begin{array}{l}=4\times \sqrt{a^2{\cos }^2\phi +b^2{\sin }^2\phi }\times \frac{ab}{\sqrt{a^2{\cos }^2\phi +b^2{\sin }^2\phi }}\\ =4ab\end{array}$

$=$constant

Hence, the result.