If the tangent to the curve $y=\frac{x}{x^{2}-3}, x \in R,(x \neq \pm \sqrt{3})$ at a point $(\alpha, \beta) \neq(0,0)$ on it is parallel to the line $2 x+6 y-11=0$ then

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Application of Derivatives

If the tangent to the curve

y = \frac{x}{x ^{2} - 3}, x \in R, (x \neq = \pm 3)

at a point

(α, β) \neq = (0, 0)

on it is parallel to the line

2 x + 6 y - 11 = 0

then

Answer: (C)

The given line

2 x + 6 y - 11 = 0

has slope

- \frac{1}{3}

, so we need the tangent slope to equal that. Differentiating

y = \frac{x}{x ^{2} - 3}

by the quotient rule gives

\frac{d y}{d x} = \frac{- ( x ^{2} + 3 )}{( x ^{2} - 3 ) ^{2}}

, and setting this to

- \frac{1}{3}

leads to

(x^{2} - 3)^{2} = 3 (x^{2} + 3)

, whose nonzero solution is

x^{2} = 9

, i.e.

α = \pm 3

. Then

β = \frac{α}{α ^{2} - 3} = \pm \frac{1}{2}

, and either way

∣6 α + 2 β ∣ = ∣ \pm (18 + 1) ∣ = 19

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option