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Question
The normal at the point (1, 1) on the curve 2y + x2 = 3 is _____________ .
Options
x + y = 0
x − y = 0
x + y + 1 = 0
x − y = 1
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Solution
`x − y = 0`
\[\text { Given }: \]
\[2y + x^2 = 3\]
\[ \Rightarrow 2\frac{dy}{dx} + 2x = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 2x}{2} = - x\]
\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) =-1\]
\[\text { Slope of the normal },m=\frac{- 1}{\text { Slope of the tangent }}=\frac{- 1}{- 1}=1\]
\[\text { Now }, \]
\[\left( x_1 , y_1 \right) = \left( 1, 1 \right)\]
\[ \therefore \text { Equation of the normal }\]
\[ = y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 1 = 1 \left( x - 1 \right)\]
\[ \Rightarrow x - y = 0\]
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