Question

Find the equation of the normal to the curve $x^2=4y$ which passes through the point $(1,2)$

Answer 1

Curve is $x^2=4y$

$\Rightarrow 2x\frac{dx}{dx}=4\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{2x}{4}=\frac{x}{2}$

Slope of the normal$=\frac{-1}{\frac{dy}{dx}}=\frac{-1}{\frac{x}{2}}=\frac{-2}{x}$

Let $(h,k)$ be the point where normal and curve intersect.

$\therefore$ Slope of normal at $(h,k)$ is $=\frac{-2}{h}$

Equation of normal passing through $(h,k)$ with slope $\frac{-2}{h}$

is $y-y_1=m(x-x_1)$

$y-k=\frac{-2}{h}(x-h)$

Since normal passes through $(1,2)$, it will satisfy its equation

$2-k=\frac{-2}{h}(1-h)$

$\Rightarrow k=2+\frac{2}{h}(1-h)$ ........$\left(1\right)$

Since $(h,k)$ lies on the curve $x^2=4y$

$h^2=4k$ or $k=\frac{h^2}{4}$ ........$\left(2\right)$

Using $\left(1\right)$ and $\left(2\right)$

$2+\frac{2}{h}(1-h)=\frac{h^2}{4}$

$\Rightarrow 2+\frac{2}{h}-2=\frac{h^2}{4}$

$\Rightarrow \frac{2}{h}=\frac{h^2}{4}$

$\Rightarrow h^3=8$

$\Rightarrow h=8^{\frac{1}{3}}$

Put $h=2$ in eqn$\left(2\right)$ we get

$k=\frac{h^2}{4}=\frac{2^2}{4}=1$

Hence $h=2,k=1$

Put $h=2$ and $k=1$ in equation of normal we get

$y-1=\frac{-2}{2}(x-2)$

$\Rightarrow y-1=-x+2$

$\therefore x+y=3$