If the curves, $$\frac{x^2}{a} + \frac{y^2}{b} = 1$$ and $$\frac{x^2}{c} + \frac{y^2}{d} = 1$$ intersect each other at an angle of 90°, then which of the following relations is TRUE?

Given curves

$$\frac{x^2}{a}+\frac{y^2}{b}=1 \quad\cdots(1)$$ and $$\frac{x^2}{c}+\frac{y^2}{d}=1 \quad\cdots(2)$$ intersect orthogonally.

We need to find the correct relation among $$a,b,c,d.$$

Differentiate (1):

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

$$\frac{dy}{dx}=-\frac{bx}{ay}$$

Hence slope of tangent to (1) is $$m_1=-\frac{bx}{ay}$$

Now differentiate (2): $$\frac{2x}{c}+\frac{2y}{d}\frac{dy}{dx}=0$$

$$\frac{dy}{dx} =-\frac{dx}{cy}$$

Hence slope of tangent to (2) is $$m_2=-\frac{dx}{cy}$$

Since the curves intersect at $$90^\circ,$$ their tangents are perpendicular.

Therefore,

$$m_1m_2=-1$$

Substituting,

$$\left(-\frac{bx}{ay}\right) \left(-\frac{dx}{cy}\right)=-1$$

$$\frac{bdx^2}{acy^2}=-1 \quad\cdots(3)$$

Now from equations (1) and (2),

$$\frac{x^2}{a}+\frac{y^2}{b}=1$$

$$\frac{x^2}{c}+\frac{y^2}{d}=1$$

Subtracting,

$$x^2\left(\frac1a-\frac1c\right)+y^2\left(\frac1b-\frac1d\right)=0$$

$$x^2\frac{c-a}{ac} +y^2\frac{d-b}{bd}=0$$

$$bd(c-a)x^2 +ac(d-b)y^2=0$$

$$bd(a-c)x^2 =ac(d-b)y^2 \quad\cdots(4)$$

Using (3),

$$\frac{x^2}{y^2} =\frac{-ac}{bd}$$

Substitute into (4):

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

$$-(a-c)=d-b$$

$$a-c=b-d$$

Hence,

$$a-b=c-d$$

Therefore, the correct option is

$$\boxed{a-b=c-d}$$