Mathematics - Parabola Question with Solution | TestHub

Let $P\left(2t,t^2\right)$.

Equation normal at $P$ to $x^2=4y$ be

$y-t^2=-\frac{1}{t}\left(x-2t\right)$.

It passes a normal through $\left(-3,0\right)$.

$\Rightarrow 0-t^2=-\frac{1}{t}\left(-3-2t\right)$

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

$\Rightarrow \left(t+1\right)\left(t^2-t+3\right)=0$

$\Rightarrow t=-1$

So, point $P$ is $\left(-2,1\right)$.

Equation of a tangent to $x^2=4y$at $\left(-2,1\right)$.

$\Rightarrow x\left(-2\right)=2\left(y+1\right)$

$\therefore x+y+1=0$.