Question

Normal at $(1, 1)$ on the curve $2y=9-x^2$ is

• A. $x + y = 0$
• B. $x - y = 0$
• C. $- x - 2y = 0$
• D. $2x -y = 0.$

Answer 1

Answer: (B)

$x - y = 0$

Answer 2

Equation of curve $2y = 9 - x^2$ $\therefore 2 \frac{dy}{dx}=-2x$ $\Rightarrow \frac{dy}{dx}=-x$ $\therefore \left|\frac{dy}{dx}\right|_{at \left(1, 1\right)}=-1$ $\therefore$ slope of the normal $=-\frac{1}{\left(\frac{dy}{dx}\right)_{\left(1,1\right)}}=1$ $\therefore$ equation of normal at $\left(1, 1\right)$ is $y - 1 = 1\left(x - 1\right)$ $\Rightarrow y-x=1-1$ $\Rightarrow y-x=0$ $\Rightarrow x-y=0$