The slope of the tangent drawn at (1, 4) to t | vedclass

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(D) Given equation: $2x \sqrt{y} + 2y \sqrt{x} = 4x \sqrt{x} + y \sqrt{y}$.Divide the entire equation by $x \sqrt{x}$ (assuming $x > 0$):$2 \frac{\sqr

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$4$

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(D) Given equation: $2x \sqrt{y} + 2y \sqrt{x} = 4x \sqrt{x} + y \sqrt{y}$.Divide the entire equation by $x \sqrt{x}$ (assuming $x > 0$):$2 \frac{\sqrt{y}}{\sqrt{x}} + 2 \frac{y}{x} = 4 + \frac{y \sqrt{y}}{x \sqrt{x}}$.Let $v = \sqrt{\frac{y}{x}}$,then $v^2 = \frac{y}{x}$.The equation becomes $2v + 2v^2 = 4 + v^3$,which is $v^3 - 2v^2 - 2v + 4 = 0$.Factoring: $v^2(v - 2) - 2(v - 2) = 0 \Rightarrow (v^2 - 2)(v - 2) = 0$.At $(1, 4)$,$v = \sqrt{\frac{4}{1}} = 2$,which satisfies the equation.For a homogeneous equation of the form $f(\frac{y}{x}) = c$,the derivative is $\frac{dy}{dx} = \frac{y}{x}$.At $(1, 4)$,the slope is $\frac{dy}{dx} = \frac{4}{1} = 4$.

The slope of the tangent drawn at $(1, 4)$ to the curve given implicitly by the equation $2(x \sqrt{y} + y \sqrt{x}) = 4x \sqrt{x} + y \sqrt{y}$ is -

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