Question

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

• A. $\frac{3 y-4 x-1}{2 y-3 x+2}$
• B. $\frac{3 y+4 x+1}{2 y+3 x+2}$
• C. $\frac{3 y-4 x+1}{2 y-3 x-2}$
• D. $\frac{3 y-4 x+1}{2 y+3 x+2}$

Answer 1

Answer: (A)

We have, $2 x^{2}-3 x y+ y^{2}+x+2 y-8=0$
On differentiating w.r.t. $x$, we get
$4 x-3 x \frac{d y}{d x}-3 y+2 y \frac{d y}{d x}+1+\frac{2 d y}{d x}-0=0$
$\Rightarrow 4 x+(-3 x+2 y+2) \frac{d y}{d x}-3 y+1=0$
$\Rightarrow \frac{d y}{d x}=\frac{3 y-4 x-1}{2 y-3 x+2}$