\[ \begin{array}{l \qquad l \qquad l} (A)\; x + y = 0 \quad &(B) \;x – y = 0 \\(C)\; x + y +1 = 0 \quad & (D)\; x – y = 0\end{array} \]

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- $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$

Step 1:

The equation of the curve $2y+x^2=3$

Differentiating with respect to $x$

$2\large\frac{dy}{dx}$$+2x=0$

$2\large\frac{dy}{dx}$$=-2x$

$\large\frac{dy}{dx}$$=-x$

$\large\frac{dy}{dx}$ at $(1,1)=-1$=Slope of tangent.

Step 2:

Since slope of normal $=\large\frac{1}{slope \;of\; tangent}$$=1$

The equation of the normal is

$y-y_1$=Slope of normal$(x-x_1)$

$y-1=1.(x-1)$ or $y-1=x-1$

$x-y=0$

Part (B) is the correct answer.

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