A) \[2x+y-6=0\]
B) \[2y+x-6=0\]
C) \[x+3y-8=0\]
D) \[3x+y-8=0\]
E) \[x+y-4=0\]
Correct Answer: A
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 \[(y-2)=-2(x-2)\] \[\Rightarrow \] \[2x+y=6\]You need to login to perform this action.
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