JEE Main 2022 · 24 Jun · Shift 2 · Mathematics Q19

← JEE Main 2022 · 24 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 19 of 90

The slope of normal at any point $(x,y)$ , $x>0$ , $y>0$ on the curve $y=y(x)$ is given by $\dfrac{x^2}{xy-x^2y^2-1}$ . If the curve passes through the point $(1,1)$ , then $e\cdot y(e)$ is equal to

\dfrac{1-\tan(1)}{1+\tan(1)}

\tan(1)

\dfrac{1+\tan(1)}{1-\tan(1)}

Answer and worked solution

Official answer: D

Slope of tangent. $\dfrac{dy}{dx}=-\dfrac{xy-x^2y^2-1}{x^2}=\dfrac{1+x^2y^2-xy}{x^2}$ .

Substitute $v=xy$ . $\dfrac{dv}{dx}=y+x\dfrac{dy}{dx}=\dfrac{xy+1+v^2-xy}x=\dfrac{1+v^2}x$ .

Integrate. $\tan^{-1}v=\log_ex+C$ ; at $(1,1)$ , $C=\frac\pi4$ .

At $x=e$ . $e\,y(e)=\tan\left(1+\frac\pi4\right)=\dfrac{1+\tan1}{1-\tan1}$ .

(D)

Topics:Formation and applications Variable separable Tangents and normals

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JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 19