### JEE Main On 16 April 2018 Question 23

##### Question: Let P be a point on the parabola $ , $ $ x^{2}=4y $ .If the distance of P from the centre of the circle, $ x^{2}+y^{2}+6x+8=0 $ is minimum, then the equation of the tangent to the parabola at P, is? [JEE Main 16-4-2018]

#### Options:

A) $ x+4y-2=0 $

B) $ x+2y=0 $

C) $ x+y+1=0 $

D) $ x-y+3=0 $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

Let $ (2t,t^{2}) $ be any point on the parabola. Centre of circle $ =(-g,-f)=(-3,0) $

For the distance between point P and centre of circle to be minimum,

line drawn from the centre of circle to point P must be normal to the parabola at P.

Slope of line joining centre of circle to point P $ P=\frac{y _2-y _1}{x _2-x _1}=\frac{t^{2}-0}{2t+3} $

Slope of tangent to parabola at $ P=\frac{dy}{dx}=\frac{x}{2}=t $ Slope of normal $ =\frac{-1}{t} $

Therefore, $ \frac{t^{2}-0}{2t+3}=\frac{-1}{t} $

$ \Rightarrow $ $ t^{3}+2t+3=0 $

$ \Rightarrow $ $ (t+1)(t^{2}-t+1)=0 $ Real roots of above equation is $ t=-1 $

Coordinate of $ P=(2t,t^{2})=(-2.1) $ Slope of tangent of parabola at $ P=\frac{dy}{dx}=\frac{x}{2}=t=-1 $

Therefore, equation of tangent is $ (y-1)=(-1)(x+2) $

$ \Rightarrow $ $ x+y+1=0 $ Hence, answer is option C.