Q103 / 160 · TS EAMCET 2007 · Mathematics

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

  • Correct answer
    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
    $\frac{3 y-4 x+1}{2 y+3 x+2}$

Solution

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

Hence, the correct answer is option: 'A'.

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