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

KEAMKEAM 2009Application of Derivatives

Solution:

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 $