Question

The normal to the curve x² = 4y passing through (1, 2) is

(a) x + y = 3
(b) x − y = 3
(c) x + y = 1
(d) x − y = 1

Answer 1

Disclaimer: None of the given options is correct.

$$\begin{gathered} \mathrm{Given}: \\ {x}^2=4y \\ \Rightarrow 2x=4\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{2x}{4}=\frac{x}{2} \\ \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{\left(1,2\right)}\text{=}\frac{1}{2} \\ \text{Slope of the normal,}\mathit{\text{m}}\text{=}\frac{-1}{\text{Slope of the tangent}}\text{=}\frac{-1}{\frac{1}{2}}\text{=-2} \\ \mathrm{Also}, \\ \left(x_1,y_1\right)=\left(1,2\right) \\ \therefore \text{Equation of the normal} \\ =y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-2=-2\left(x-1\right) \\ \Rightarrow y-2=-2x+2 \\ \Rightarrow 2x+y=4 \end{gathered}$$