Question

The equation of the tangent to the curve $x^2-2xy+y^2+2x+y-6=0$ at $(2,2)$ is

• A. $2x+y-6=0$
• B. $2y+x-6=0$
• C. $x+3y-8=0$
• D. $3x+y-8=0$
• E. $x+y-4=0$

Answer 1

Answer: (A)

Given, $x^2-2xy+y^2+2x+y-6=0$
On differentiating w.r.t. $x,$ we get $2x-2\left(y+x\frac{dy}{dx}\right)+2y\frac{dy}{dx}+2\frac{dy}{dx}=0$ At (2, 2), $4-2\left(2+2\frac{dy}{dx}\right)+4\frac{dy}{dx}+2+\frac{dy}{dx}=0$
$\Rightarrow$ $\frac{dy}{dx}=-2$
$\therefore$ Equation of tangent at
(2, 2) is $(y-2)=-2(x-2)$
$\Rightarrow$ $2x+y=6$