JEE Advanced 2018 · Paper 1 · Mathematics Q16

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MathematicsSingle correct+3 / -1Question 52 of 54

Passage

Let $S$ be the circle in the $xy$ -plane defined by the equation $x^2 + y^2 = 4$ .

Let $P$ be a point on the circle $S$ with both coordinates being positive. Let the tangent to $S$ at $P$ intersect the coordinate axes at the points $M$ and $N$ . Then, the mid-point of the line segment $MN$ must lie on the curve

(x + y)^2 = 3xy

x^{2/3} + y^{2/3} = 2^{4/3}

x^2 + y^2 = 2xy

x^2 + y^2 = x^2y^2

Answer and worked solution

Official answer: D

Tangent. Let $P=(2\cos\theta,2\sin\theta)$ . The tangent is $x\cos\theta+y\sin\theta=2$ .

Intercepts. $M=\left(\dfrac2{\cos\theta},0\right)$ and $N=\left(0,\dfrac2{\sin\theta}\right)$ .

Mid-point. $(x,y)=(\sec\theta,\csc\theta)$ , so
$\frac1{x^2}+\frac1{y^2}=\cos^2\theta+\sin^2\theta=1\implies x^2+y^2=x^2y^2.$

(D)

Topics:Tangents and normals Locus

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JEE Advanced 2018 · Paper 1, Mathematics question 16