Q112 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Mathematics

Find \(\frac{d y}{d x}\) if \(2 x^2-3 x y+y^2+x+2 y-8=0\)

  • 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

Given equation \(2 x^2-3 x y+y^2+x+2 y-8=0\) On differentiating w.r.t. \(x\), we get \(\begin{aligned} & 4 x-3 x \frac{d y}{d x}-3 y+2 y \frac{d y}{d x}+1+2 \frac{d y}{d x}=0 \\ \Rightarrow & \quad(3 x-2 y-2) \frac{d y}{d x}=4 x-3 y+1 \\ \Rightarrow & \frac{d y}{d x}=\frac{4 x-3 y+1}{3 x-2 y-2}=\frac{3 y-4 x-1}{2 y-3 x+2} \end{aligned}\) Hence, options (a) is correct.

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

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