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Question
The normal to the curve x2 = 4y passing through (1, 2) is _____________ .
Options
x + y = 3
x − y = 3
x + y = 1
x − y = 1
none of these
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Solution
\[\text { Given }: \]
\[ x^2 = 4y\]
\[ \Rightarrow 2x = 4\frac{dy}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}\]
\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( 1, 2 \right) =\frac{1}{2}\]
\[\text { Slope of the normal,}m=\frac{- 1}{\text{ Slope of the tangent }}=\frac{- 1}{\frac{1}{2}}=-2\]
\[\text { Also }, \]
\[\left( x_1 , y_1 \right) = \left( 1, 2 \right)\]
\[ \therefore \text { Equation of the normal }\]
\[ = y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 2 = - 2 \left( x - 1 \right)\]
\[ \Rightarrow y - 2 = - 2x + 2\]
\[ \Rightarrow 2x + y = 4\]
Notes
None of the given options is correct.
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