If ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0,the | vedclass

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(A) Given the equation: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$Differentiating both sides with respect to $x$:$\frac{d}{dx}(ax^2) + \frac{d}{dx}(2hxy

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$-\frac{ax + hy + g}{hx + by + f}$

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(A) Given the equation: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$Differentiating both sides with respect to $x$:$\frac{d}{dx}(ax^2) + \frac{d}{dx}(2hxy) + \frac{d}{dx}(by^2) + \frac{d}{dx}(2gx) + \frac{d}{dx}(2fy) + \frac{d}{dx}(c) = 0$Applying the chain rule and product rule:$2ax + 2h(y + x\frac{dy}{dx}) + 2by\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0$Divide the entire equation by $2$:$ax + hy + hx\frac{dy}{dx} + by\frac{dy}{dx} + g + f\frac{dy}{dx} = 0$Grouping the terms containing $\frac{dy}{dx}$:$\frac{dy}{dx}(hx + by + f) = -(ax + hy + g)$Solving for $\frac{dy}{dx}$:$\frac{dy}{dx} = -\frac{ax + hy + g}{hx + by + f}$

If $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,then $\frac{dy}{dx} = $

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