Question:
The normal at the point $(1,1)$ on the curve $2 y+x^{2}=3$ is
A. $x+y=0$
B. $x-y=0$
C. $x+y+1=0$
D. $x-y=1$
Solution:
Given that the curve $2 y+x^{2}=3$ has a normal passing through point $(1,1)$.
Differentiating both the sides w.r.t. $x$,
$2 \frac{d y}{d x}+2 x=0$
Slope of the tangent $\frac{d y}{d x}=-x$
For $(1,1)$ :
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-1$
Equation of the normal:
$\left(\mathrm{y}-\mathrm{y}_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(\mathrm{x}-\mathrm{x}_{1}\right)$
$\Rightarrow(\mathrm{y}-1)=\frac{-1}{-1}(\mathrm{x}-1)$
$\Rightarrow y-1=x-1$
$\Rightarrow y-x=0$
$\Rightarrow x-y=0$
Hence, option B is correct.