Question

The normal at the point (1, 1) on the curve 2y + x² = 3 is

(a) x + y = 0
(b) x − y = 0
(c) x + y + 1 = 0
(d) x − y = 1

Answer 1

(b) x − y = 0

$$\begin{gathered} \mathrm{Given}: \\ 2y+x^2=3 \\ \Rightarrow 2\frac{dy}{dx}+2x=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{2}=-x \\ \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}\text{=-1} \\ \text{Slope of the normal,}\mathit{\text{m}}\text{=}\frac{-1}{\text{Slope of the tangent}}\text{=}\frac{-1}{-1}\text{=1} \\ \mathrm{Now}, \\ \left(x_1,y_1\right)=\left(1,1\right) \\ \therefore \text{Equation of the normal} \\ =y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-1=1\left(x-1\right) \\ \Rightarrow x-y=0 \end{gathered}$$