Given curves, $\frac{x^2}{a}+\frac{y^2}{b}=1 .....\left(i\right)$

On differentiating equation $\left(i\right)$, we get

$\frac{2x}{a}+\frac{2y}{b}\frac{dy}{dx}=0$

$\Rightarrow m_1=\frac{dy}{dx}=-\frac{xb}{ya} ...\left(iii\right)$

and $\frac{x^2}{c}+\frac{y^2}{d}=1.....\left(ii\right)$ 

On differentiating equation $\left(ii\right)$, we get

$\frac{2x}{c}+\frac{2y}{d}\frac{dy}{dx}=0$

$\Rightarrow m_2=\frac{dy}{dx}=-\frac{xd}{yc} ...\left(iv\right)$

For orthogonal curves, $m_1\times m_2=-1$

$\Rightarrow \left(-\frac{xb}{ya}\right)\times \left(-\frac{xd}{yc}\right)=-1$

$\frac{x^2}{y^2}=-\frac{ac}{bd}....\left(v\right)$

From equation $\left(i\right)-\left(ii\right)$, we get

$x^2\left(\frac{c-a}{ac}\right)=y^2\left(\frac{b-d}{bd}\right)$

$\Rightarrow \frac{x^2}{y^2}=\left(\frac{b-d}{c-a}\right)\left(\frac{ac}{bd}\right)$

From equation $\left(v\right)$, we get

 $a-c=b-d$

$\Rightarrow a-b=c-d$.