Question

The normal to the curve $x^2=4y$ passing through $(1,2)$ is

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

Answer 1

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

Curve is $x^2=4y$

Diff wrt to x

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

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

Let (h, k) be the point where normal and curve intersects

$\therefore$ Slope of normal at (h, k)$=-2/h$

Equation of normal passes through (h, k)

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

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

Since normal passes through $(1,2)$

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

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

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

$h^2=4k$

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

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

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

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

$h=2$

Putting $h=2$ in $\left(2\right)$

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

$h=2$ and $k=1$ putting in equation of normal

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

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

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

$\Rightarrow x+y=2+1$

$\Rightarrow x+y=3$.