Let a curve $y=f(x), x \in(0, \infty)$ pass through the points $P\left(1, \frac{3}{2}\right)$ and $Q\left(a, \frac{1}{2}\right)$. If the tangent at any point $R(b, f(b))$ to the given curve cuts the y-axis at the point $S(0, c)$ such that $bc=3$, then $(PQ)^{2}$ is equal to ____.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Differential Equations

Let a curve

y = f (x), x \in (0, \infty)

pass through the points

P (1, \frac{3}{2})

and

Q (a, \frac{1}{2})

. If the tangent at any point

R (b, f (b))

to the given curve cuts the y-axis at the point

S (0, c)

such that

b c = 3

, then

(P Q)^{2}

is equal to ____.

Answer: 5

The tangent at

R (b, f (b))

has equation:

y - f (b) = f^{'} (b) (x - b)

S (0, c)

c - f (b) = - b f^{'} (b)

Given

b c = 3

, so

c = \frac{3}{b}

\frac{3}{b} - f (b) = - b f^{'} (b)

b f^{'} (b) - f (b) = - \frac{3}{b}

Dividing by

b^{2}

\frac{b f ^{'} ( b ) - f ( b )}{b ^{2}} = - \frac{3}{b ^{3}}

This is

\frac{d}{d b} (\frac{f ( b )}{b}) = - \frac{3}{b ^{3}}

Integrating:

\frac{f ( b )}{b} = \frac{3}{2 b ^{2}} + λ

Using

P (1, \frac{3}{2})

\frac{3}{2} = \frac{3}{2} + λ \Rightarrow λ = 0

f (b) = \frac{3}{2 b}

For point

Q (a, \frac{1}{2})

\frac{1}{2} = \frac{3}{2 a} \Rightarrow a = 3

Therefore

Q (3, \frac{1}{2})

and

(P Q)^{2} = (3 - 1)^{2} + (\frac{1}{2} - \frac{3}{2})^{2} = 4 + 1 = 5

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer