Question

If $2x^2-3xy+y^2+x-2y-8=0,$ 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, $2x^2-3xy+y^2+x-2y-8=0$
On differentiating w.r.t.x, we get
$4x-3\left(x\frac{dy}{dx}+y\right)+2y\frac{dy}{dx}+1-2\frac{dy}{dx}=0$
$\Rightarrow$ $4x-3x\frac{dy}{dx}-3y+2y\frac{dy}{dx}-2\frac{dy}{dx}=-1$
$\Rightarrow$ $\frac{dy}{dx}(-3x+2y-2)=-1-4x+3y$
$\Rightarrow$ $\frac{dy}{dx}=\frac{3y-4x-1}{2y-3x-2}$