The equation of the normal to the curve $2 y = 3 - x^{2}$ at $(1, 1)$ is -

x + y = 0

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x + y + 1 = 0

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x - y + 1 = 0

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x - y = 0

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Solution

The correct option is D $x - y = 0$
$2 y = 3 - x^{2}$
$\Rightarrow 2 \frac{d y}{d x} = - 2 x \Rightarrow \frac{d y}{d x} = - x$
$\Rightarrow∴ {(\frac{d y}{d x})}_{(1, 1)} = - 1 = m$ (say)
$∴$ Slope of normal at the given point is $= - \frac{1}{m} = 1$
Therefore, equation of normal is, $(y - 1) = 1 (x - 1)$
$\Rightarrow x - y = 0$
Hence, option 'D' is correct.

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Q. 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

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Choose the correct answer in the following question:

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The equation of the normal to the curve 2y=3−x2 at (1,1) is -

The correct option is D x−y=02y=3−x2⇒2dydx=−2x⇒dydx=−x⇒∴(dydx)(1,1)=−1=m (say)∴ Slope of normal at the given point is =−1m=1Therefore, equation of normal is, (y−1)=1(x−1)⇒x−y=0Hence, option 'D' is correct.

The equation of the normal to the curve $2 y = 3 - x^{2}$ at $(1, 1)$ is -