If the curves, x ^ 2 a + y ^ 2 b = 1 and x ^ 2 c + y ^ 2 d = 1...

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

On differentiating equation \left(i\right)$, we get</p><p>$\dfrac{2x}{a} + \dfrac{2y}{b}\dfrac{dy}{dx} = 0$</p><p>$⇒ {m}_{1} = \dfrac{dy}{dx} = - \dfrac{xb}{ya} . . . \left(iii\right)$</p><p>and$\dfrac{{x}^{2}}{c} + \dfrac{{y}^{2}}{d} = 1 . . . . . \left(ii\right)$</p><p>On differentiating equation \left(ii\right)$, we get

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

$⇒ {m}_{2} = \dfrac{dy}{dx} = - \dfrac{xd}{yc} . . . \left(iv\right)$

For orthogonal curves, ${m}_{1} \times {m}_{2} = - 1$

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

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

From equation \left(i\right) - \left(ii\right)$, we get</p><p>${x}^{2}\left(\dfrac{c - a}{ac}\right) = {y}^{2}\left(\dfrac{b - d}{bd}\right)$</p><p>$⇒ \dfrac{{x}^{2}}{{y}^{2}} = \left(\dfrac{b - d}{c - a}\right)\left(\dfrac{ac}{bd}\right)$</p><p>From equation \left(v\right)$, we get

$a - c = b - d$

$⇒ a - b = c - d$.