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Let P be the point on parabola y^2=4x wh...

Let P be the point on 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

A

(a) `SP=2sqrt(5)`

B

(b) `SQ:QP=(sqrt(5)+1):2`

C

(c) the x-intercept of the normal to the parabola at P is 6

D

(d) the slope of the tangent to the circle at Q is `(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
1,3,4
Given circle is :
`x^(2)+y^(2)-4x-16y+64=0`
`"Or "(x-2)^(2)+(y-8)^(2)=4`
Centre `S-=(2,8)`
Normal to parabola is `y=mx-2m-m^(3)`

For shortest diatance from the centre, normal must pass through the centreof the circle.
`rArr8=2m-2m-m^(3)`
`rArrm=-2`
`rArr" Normal at"P,y=-2x+12`
`:." Point"P-=(am^(2),-2am)-=(4,4)`
`:." "SP=sqrt((4-2)^(2)+(8-8)^(2))=2sqrt(5)`
`:." "SQ:QP=2:(2sqrt(5)-2)`
Slope of tangent at `Q=(1)/(2)` (Tangent at Q is perpendicular to SP)
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