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The equation of normal to the curve $2 y+x^2=3$ at the point $(1,1)$ is: |
$x+y=0$ $x-y=0$ $x+y+1=0$ $x-y=1$ |
$x-y=0$ |
The correct answer is Option (2) → $x-y=0$ $2 y+x^2=3$ differentiating w.r.t. x $⇒\frac{2dy}{dx}+2x=3⇒\frac{dy}{dx}=-x$ ⇒ Slope of normal = $\frac{1}{x}$ $\left.\frac{1}{x}\right]_{(1,1)}=1$ so equation of normal $y-1=x-1$ so $x-y=0$ |
