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

Answer 1

\[\text { Given }: \]

\[ x^2 = 4y\]

\[ \Rightarrow 2x = 4\frac{dy}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}\]

\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( 1, 2 \right) =\frac{1}{2}\]

\[\text { Slope of the normal,}m=\frac{- 1}{\text{ Slope of the tangent }}=\frac{- 1}{\frac{1}{2}}=-2\]

\[\text { Also }, \]

\[\left( x_1 , y_1 \right) = \left( 1, 2 \right)\]

\[ \therefore \text { Equation of the normal }\]

\[ = y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - 2 = - 2 \left( x - 1 \right)\]

\[ \Rightarrow y - 2 = - 2x + 2\]

\[ \Rightarrow 2x + y = 4\]