CUET Preparation Today
The normal at the point (1, 1) on the curve 2y = 3 - x2 is |
x + y = 0 x + y + 1 = 0 x – y + 1 = 0 x – y = 0 |
x – y = 0 |
$2 y=3-x^2 \Rightarrow 2 \frac{d y}{d x}=-2 x \Rightarrow \frac{d y}{d x}=-x$ ⇒ slope of normal $=\frac{1}{x}$ ∴ slope of normal at $(1,1)=\frac{1}{1}=1$ Equation of normal is $y-1=1(x-1) \Rightarrow x-y=0$ |
