Question

The equation of normal to the parabola $x^2=4y$ drawn from the point $(1,2)$ is

• A. $2x+y=4$
• B. $x+y=3$
• C. $x+3y=7$
• D. $x-y=1$

Answer 1

The correct option is B $x+y=3$

Let point $Q$ $(h,\frac{h^2}{4})$ be the foot of the normal.
Let $P$ be $(1,2)$
Now, $m_{PQ}=$ slope of the normal at $Q\cdots \left(i\right)$
$x^2=4y$
Differentiating w.r.t. $x$,
$\Rightarrow 2x=4\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{x}{2}$
$\Rightarrow \text{Slope of the normal at}Q=-{\left(\frac{dx}{dy}\right)}_{x=h}=-\frac{2}{h}$
Using $\left(i\right)$, we get
$\Rightarrow \frac{\frac{h^2}{4}-2}{h-1}=-\frac{2}{h}$
$\Rightarrow \frac{h^3}{4}-2h=-2h+2$
$\Rightarrow h^3=8\Rightarrow h=2$
Hence, the co-ordinates of point $Q$ is $(2,1)$, and so the equation of the required normal becomes $x+y=3$.