Question

Sum of slopes of tangent and normal to the curve $x^2+y^2+2x+2y=6$ at $(1,1)$ is

• A. $\frac{2}{3}$
• B. $-1$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

Given, $x^2+y^2+2x+2y=6$
Differentiating w.r.t. $x$
$\Rightarrow 2x+2y\frac{dy}{dx}+2+2\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\left(\frac{x+1}{y+1}\right)$
$\Rightarrow {\frac{dy}{dx}|}_{x=1}=-1$
$\Rightarrow$ Slope of tangent $=-1$
$\Rightarrow$ Slope of normal $=1$
$\therefore$ Sum of Slopes = $-1+1=0$