Question

If $ 2{{x}^{2}}-3xy+{{y}^{2}}+x-2y-8=0, $ then $ \frac{dy}{dx} $ is equal to

• 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} $

Answer 1

Answer: (A)

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} $