Question:
The normal to the curve $x^{2}=4 y$ passing through $(1,2)$ is
A. $2 x+y=4$
B. $x-y=3$
C. $x+y=1$
D. $x-y=1$
Solution:
Given that the curve $x^{2}=4 y$
Differentiating both the sides w.r.t. $x$,
$4 \frac{d y}{d x}=2 x$
Slope of the tangent $\frac{d y}{d x}=\frac{1}{2} x$
For $(1,2)$ :
$\Rightarrow \frac{d y}{d x}=\frac{1}{2}$
Equation of the normal:
$\left(\mathrm{y}-\mathrm{y}_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(\mathrm{x}-\mathrm{x}_{1}\right)$
$\Rightarrow(y-2)=\frac{-2}{1}(x-1)$
$\Rightarrow y-2=-2 x+2$
$\Rightarrow y+2 x=4$
No option matches the answer.