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JEE Advanced · Mathematics · 13. Parabola

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 -

  1. A SP=25
  2. B SQ :QP=5+1 :2
  3. C The x - intercept of the normal to the parabola at P is 6
  4. D The slope of the tangent to the circle at Q is 12
Verified Solution

Answer & Solution

Correct Answer

(A) SP=25

Step-by-step Solution

Detailed explanation


y2=4x
Point P lies on normal to parabola passing through centre of circle
y+tx=2t+t3 ......(i)
8+2t=2t+t3
t=2
P4, 4
SP= 4-22+4-82
SP=25
SQ=2
PQ=25-2
SQQP=15-1=5+14
To find x intercept
Put y=0 in (i)
x=2+t2
x=6
Slope of common normal = -t= -2
Slope of tangent =12
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