JEE Advanced 2016 Paper 2

Section: Mathematics

Question : 47 of 54

Marks: +1, -0

Let P be the point on the parabola y2=4x which is at the shortest distance from the center S of the circle x2+y2–4x–16y+64=0. Let Q be the point on the circle dividing the line segment SP internally. Then-

[JEE Adv 2016 P2]

SP=2√5
SQ:QP=(√5+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 is
1
2

Validate

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