JEE Advanced 2015 · Paper 2 · Mathematics Q52

← JEE Advanced 2015 · Paper 2

MathematicsMultiple correct+4 / -2Question 52 of 60

Consider the hyperbola $H:x^2-y^2=1$ and a circle $S$ with center $N(x_2,0)$ . Suppose that $H$ and $S$ touch each other at a point $P(x_1,y_1)$ with $x_1>1$ and $y_1>0$ . The common tangent to $H$ and $S$ at $P$ intersects the $x$ -axis at point $M$ . If $(l,m)$ is the centroid of the triangle $\Delta PMN$ , then the correct expression(s) is(are)

\dfrac{dl}{dx_1}=1-\dfrac1{3x_1^2}

for

x_1>1

\dfrac{dm}{dx_1}=\dfrac{x_1}{3\left(\sqrt{x_1^2-1}\right)}

for

x_1>1

\dfrac{dl}{dx_1}=1+\dfrac1{3x_1^2}

for

x_1>1

\dfrac{dm}{dy_1}=\dfrac13

for

y_1>0

Answer and worked solution

Official answer: A, B, D

Tangent. At P it is $xx_1-yy_1=1$ , so $M=\left(\frac1{x_1},0\right)$ .

Normal. It passes through the circle's centre N. The normal to the hyperbola at P meets the $x$ -axis at $x=2x_1$ , so $N=(2x_1,0)$ .

Centroid.
$l=\frac{x_1+\frac1{x_1}+2x_1}3=x_1+\frac1{3x_1},\qquad m=\frac{y_1}3.$

Derivatives.

(A) $\dfrac{dl}{dx_1}=1-\dfrac1{3x_1^2}$ . True.
(B) With $y_1=\sqrt{x_1^2-1}$ : $\dfrac{dm}{dx_1}=\dfrac{x_1}{3\sqrt{x_1^2-1}}$ . True.
(D) $\dfrac{dm}{dy_1}=\dfrac13$ . True.
(C) False.

(A), (B), (D)

Topics:Hyperbola Tangents and normals to conics Chain rule and implicit differentiation

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JEE Advanced 2015 · Paper 2, Mathematics question 52