JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Application of Derivatives

An angle of intersection of the curves,

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

and

x^{2} + y^{2} = ab, a > b

is:

Answer: (B)

Solution

Substituting

y^{2} = ab - x^{2}

into the ellipse equation gives the intersection with

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

and

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

, so

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

. The slopes there are

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

for the ellipse and

m_{2} = - \frac{x}{y}

for the circle. Then

tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} = \frac{\frac{x}{y} \cdot \frac{a ^{2} - b ^{2}}{a ^{2}}}{1 + \frac{b ^{2}}{a ^{2}} \cdot \frac{x ^{2}}{y ^{2}}}

. With

\frac{x}{y} = \frac{a}{b}

and

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

, this reduces to

\frac{a - b}{ab}

. The angle of intersection is

tan^{- 1} \frac{a - b}{ab}

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option