The normal to the curve
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.

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