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Let P be a point on parabola x ^ 2 ...

Let P be a point on parabola ` x ^ 2 = 4y ` . If the distance of P from the centre of circle ` x ^ 2 + y ^ 2 + 6x + 8 = 0 ` is minimum, then the equation of tangent at P on parabola ` x ^ 2 = 4y ` is :

A

`x+y+1=0`

B

`x+y-1=0`

C

`x-y+1=0`

D

`x-y-1=0`

Text Solution

Verified by Experts

The correct Answer is:
A
`x^2 =4y " " x^2+y^2+6x+8=0`

Line through P and C must be normal
`P(2t,t^2), C(-3,0)`
`x+ty=t^3+2t rArr -3 + 0 = t^3 +2t`
`rArr t^3 + 2t +3 = 0 rArr t=-1 rArr` P(-2,1)
Tangent at P :
`-2x=2(y+1) rArr x+y+1=0`
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