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Question

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


A
x+y+1=0
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B
x+y=0
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C
xy+1=0
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D
xy=0
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Solution

The correct option is D $$x-y=0$$
The curve is $$2y=3-{x}^{2}$$

Slope$$=2{y}^{\prime}=-2x$$ or $$\dfrac{dy}{dx}=-x$$

Slope at $$\left(1,1\right)$$ is $$\dfrac{dy}{dx}=-1$$

Equation of normal is $$y-1=\dfrac{-1}{-1}\left(x-1\right)$$

$$\Rightarrow x-1-y+1=0$$

$$\Rightarrow x-y=0$$ is the equation of the normal.

Mathematics

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