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 :

• A. $x+y+1=0$
• B. $x+4y-2=0$
• C. $x+2y=0$
• D. $x-y+3=0$

Answer 1

The correct option is A $x+y+1=0$

$\text{Let}P(2t,t^2)\text{Centre of the circle :}(-3,0)y=D^2=(2t+3{)}^2+(t^2-0{)}^2\text{For minimum distance,}\frac{dy}{dt}=0\Rightarrow 2(2t+3)2+4t^3=0\Rightarrow t^3+2t+3=0\Rightarrow (t+1)(t^2-t+3)=0\Rightarrow t=-1\therefore P\text{is}(-2,1)\text{Equation of tangent to the parabola is,}x{x}_1=2a(y+y_1)\Rightarrow x(-2)=2\times 1(y+1)\Rightarrow x+y+1=0$