If the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ intersect orthogonally, then 

$a^2-b^2=c^2-d^2$

The equations of the two curves are

$C_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$         .....(i)

$C_2: \frac{x^2}{c^2}+\frac{y^2}{d^2}=1$         .....(ii)

Suppose these two curves intersect at $P\left(x_1, y_1\right)$. Then,

$\frac{x_1{ }^2}{a^2}+\frac{y_1{ }^2}{b^2}=1$         .....(iii)

and, $\frac{x_1{ }^2}{c^2}+\frac{y_1{ }^2}{d^2}=1$         .....(iv)

Differentiating (i) and (ii) with respect to x, we get

$\left(\frac{d y}{d x}\right)_{C_1}=-\frac{b^2 x}{a^2 y}$ and $\left(\frac{d y}{d x}\right)_{C_2}=-\frac{d^2 x}{c^2 y}$

If the two curves intersect orthogonally at P. Then,

$-\frac{b^2 x_1}{a^2 y_1} \times-\frac{d^2 x_1}{c^2 y_1}=-1 \Rightarrow b^2 d^2 x_1{ }^2=-a^2 c^2 y_1{ }^2$         .....(v)

Subtracting (iv) from (iii), we get

$\frac{x_1{ }^2}{a^2}-\frac{x_1{ }^2}{c^2}+\frac{y_1{ }^2}{b^2}-\frac{y_1{ }^2}{d^2}=0$

$\Rightarrow x_1{ }^2 \frac{\left(a^2-c^2\right)}{a^2 c^2}=y_1{ }^2 \frac{\left(d^2-b^2\right)}{b^2 d^2}$

$\Rightarrow b^2 d^2 x_1{ }^2\left(a^2-c^2\right)=a^2 c^2 y_1{ }^2\left(d^2-b^2\right)$

$\Rightarrow a^2-c^2=-\left(d^2-b^2\right)$                  [Using (v)]

$\Rightarrow a^2-b^2=c^2-d^2$