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  • If ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0, then show that dy/dx.x/dy = 1

If ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0, then show that \frac{d y}{d x} \cdot \frac{d x}{d y}=1

Answers (1)

Given: ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0

Differentiating the above with respect to x, we get

\\ 2 \mathrm{ax}+2 \mathrm{~h}\left[\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\right]+\mathrm{b} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}^{2}\right)+2 \mathrm{~g}+2 \mathrm{f} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{y}=0 \\ \Rightarrow 2 \mathrm{ax}+2 \mathrm{hx} \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{hy}+2 \mathrm{by} \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{~g}+2 \mathrm{f} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}[2 \mathrm{hx}+2 \mathrm{by}+2 \mathrm{f}]=-2 \mathrm{ax}-2 \mathrm{hy}-2 \mathrm{~g} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-[2 \mathrm{ax}+2 \mathrm{hy}+2 \mathrm{~g}]}{2 \mathrm{hx}+2 \mathrm{by}+2 \mathrm{f}} \ldots(\mathrm{ii})

Now, we again differentiate eq (i) with respect to y, we get,

\\ a \frac{d}{d y}\left(x^{2}\right)+2 h\left[x+y \frac{d}{d y} x\right]+2 y b+2 g \frac{d x}{d y}+2 f=0 \\ \Rightarrow 2 a x \frac{d x}{d y}+2 h x+2 h y \frac{d x}{d y}+2 b y+2 g \frac{d x}{d y}+2 f=0 \\ \Rightarrow \frac{d x}{d y}[2 a x+2 h y+2 g]=-2 h x-2 b y-2 f \\ \Rightarrow \frac{d x}{d y}=\frac{-[2 h x+2 b y+2 f]}{2 a x+2 h y+2 g} \ldots(i i i)

Now, multiplying Eq. (ii) and (iii), we get

\\ \Rightarrow \frac{d y}{d x} \times \frac{d x}{d y}=\frac{-[2 a x+2 h y+2 g]}{2 h x+2 b y+2 f} \times \frac{-[2 h x+2 b y+2 f]}{2 a x+2 h y+2 g} =1

Hence Proved

Posted by

infoexpert22

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