JEE Advanced 2016 · Paper 2 · Mathematics Q48

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MathematicsMultiple correct+4 / -2Question 48 of 54

Let $P$ be the point on the parabola $y^2=4x$ which is at the shortest distance from the center $S$ of the circle $x^2+y^2-4x-16y+64=0$ . Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then

SP=2\sqrt5

SQ:QP=(\sqrt5+1):2

the

x

-intercept of the normal to the parabola at

P

is 6

the slope of the tangent to the circle at

Q

\dfrac12

Answer and worked solution

Official answer: A, C, D

Circle. Centre $S=(2,8)$ , radius 2.

Nearest point. The shortest distance runs along a normal to the parabola. The normal $y=mx-2m-m^3$ passes through $(2,8)$ when $m=-2$ . So $P=(4,4)$ .

(A) $SP=\sqrt{4+16}=2\sqrt5$ . True.
(B) $SQ=2$ and $QP=2\sqrt5-2$ , so $SQ:QP=(\sqrt5+1):4$ . False.
(C) The normal at P is $y=-2x+12$ , with $x$ -intercept 6. True.
(D) SP has slope −2, so the tangent at Q (perpendicular to SQ) has slope $\frac12$ . True.

(A), (C), (D)

Topics:Parabola Tangents and normals Tangents and normals to conics

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JEE Advanced 2016 · Paper 2, Mathematics question 48