A) \[\frac{3y-4x-1}{2y-3x-2}\]
B) \[\frac{3y+4x-1}{2y+3x+2}\]
C) \[\frac{3y+4x+1}{2y-3x+2}\]
D) \[\frac{3y+4x+1}{2y-3x-2}\]
Correct Answer: A
Solution :
Given, \[2{{x}^{2}}-3xy+{{y}^{2}}+x-2y-8=0\] On differentiating w.r.t.x, we get \[4x-3\left( x\frac{dy}{dx}+y \right)+2y\frac{dy}{dx}+1-2\frac{dy}{dx}=0\] \[\Rightarrow \] \[4x-3x\frac{dy}{dx}-3y+2y\frac{dy}{dx}-2\frac{dy}{dx}=-1\] \[\Rightarrow \] \[\frac{dy}{dx}(-3x+2y-2)=-1-4x+3y\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{3y-4x-1}{2y-3x-2}\]
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