Animated Solution for Mathematics - Differentiation: 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 $90^{\circ }$, then which of the following relations is true?

Welcome, future engineer. Today, we are not just solving a problem; we are witnessing a beautiful, silent dance between two mathematical curves. Imagine two paths crossing each other in the coordinate plane.

Usually, they might cross at any arbitrary angle, but today, we are dealing with a special case: orthogonality. When two curves intersect at $90^{\circ }$, they are locked in a perfect perpendicular embrace.

Our mission is to uncover the hidden relationship between the constants $a,b,c,$ and $d$ that govern these curves. Let us begin by looking at our two protagonists:

The Calculus of Slopes

To understand how these curves behave at their intersection point $P(x,y)$, we need to know their direction. In calculus, direction is synonymous with the slope of the tangent.

Let us take the first curve, $\frac{x^2}{a}+\frac{y^2}{b}=1$, and differentiate it with respect to $x$. Applying the power rule and the chain rule, we get:

With a quick algebraic shuffle, we isolate the slope $m_1$:

Now, we repeat this elegant process for the second curve, $\frac{x^2}{c}+\frac{y^2}{d}=1$. The derivative yields $\frac{2x}{c}+\frac{2y}{d}\frac{dy}{dx}=0$, which gives us our second slope:

The Orthogonality Condition

Now, we invoke the condition of orthogonality. If these curves meet at $90^{\circ }$, their tangents must be perpendicular. The condition for perpendicularity is simple yet profound: $m_1\cdot m_2=-1$.

Let us substitute our expressions:

The negative signs vanish, leaving us with $\frac{bd{x}^2}{ac{y}^2}=-1$. By cross-multiplying, we arrive at a vital intermediate relation:

The Algebraic Alchemy

We have the slopes and the condition, but we still have $x$ and $y$ cluttering our equation. We need to eliminate them by using the original equations themselves.

Subtracting the second equation from the first, the constants on the right side cancel out:

This leaves us with:

Simplifying the fractions, we obtain:

Now, we substitute our orthogonality condition $\frac{x^2}{ac}=-\frac{y^2}{bd}$ into the equation above. The expression becomes:

Since $y$ is not zero at the intersection, we can divide the entire equation by $\frac{y^2}{bd}$. We are left with:

Expanding this, we get $-c+a+d-b=0$. Rearranging the terms, we arrive at the final, elegant result:

You have just navigated the intersection of geometry and algebra. This is the essence of JEE Advanced—not just calculating, but seeing the structure beneath the surface.