If $2 x^{2} + 3 x y - y^{2} + 4 x - 5 y + 6 = 0$ , then the value of $\frac{d y}{d x}$ at (x,y)=(1,-2) is

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7/2

answer is B.

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Detailed Solution

$2 x^{2} + 3 x y - y^{2} + 4 x - 5 y + 6 = 0$
On differentiating both sides w.r.t x we get
$\begin{array}{l} 4 x + 3 y + 3 x \frac{d y}{d x} - 2 y \frac{d y}{d x} + 4 - \frac{5 d y}{d x} = 0 \\ \Rightarrow - 3 x \frac{d y}{d x} + 2 y \frac{d y}{d x} + 5 \frac{d y}{d x} = 4 x + 3 y + 4 \\ \Rightarrow (- 3 x + 2 y + 5) \frac{d y}{d x} = 4 x + 3 y + 4 \\ \Rightarrow \frac{d y}{d x} = \frac{4 x + 3 y + 4}{- 3 x + 2 y + 5} \\ At (1, - 2) \\ {(\frac{d y}{d x})}_{(1, - 2)} = \frac{4 \times 1 + 3 \times - 2 + 4}{- 3 \times 1 + 2 \times - 2 + 5} \\ = \frac{4 - 6 + 4}{- 3 - 4 + 5} = \frac{2}{- 2} = - 1 \end{array}$