JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Application of Derivatives

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

9 0^{\circ}

, then which of the following relations is TRUE ?

Answer: (B)

Solution

Differentiating

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

gives the slope

- \frac{b x}{a y}

, and the second curve gives

- \frac{d x}{cy}

. Orthogonality at a common point

(x, y)

needs the product of these slopes to be

- 1

, so

b d x^{2} = - a c y^{2}

. Subtracting the two curve equations at that point,

x^{2} (\frac{1}{a} - \frac{1}{c}) + y^{2} (\frac{1}{b} - \frac{1}{d}) = 0, \frac{x ^{2} ( c - a )}{a c} + \frac{y ^{2} ( d - b )}{b d} = 0.

Substituting

x^{2} = - \frac{a c y ^{2}}{b d}

reduces this to

\frac{y ^{2}}{b d} [(d - b) - (c - a)] = 0

, so

d - b = c - a

, that is

a - b = c - d

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option