If theta is the angle between the curves xy=2 | vedclass

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(D) The equations of the curves are:$xy=2$ $\ldots$ $(i)$$x^2+4y=0$ $\ldots$ $(ii)$From $(i)$,$y = \frac{2}{x}$. Substituting this into $(ii)$:$x^2 +

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$3$

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(D) The equations of the curves are:$xy=2$ $\ldots$ $(i)$$x^2+4y=0$ $\ldots$ $(ii)$From $(i)$,$y = \frac{2}{x}$. Substituting this into $(ii)$:$x^2 + 4(\frac{2}{x}) = 0 \Rightarrow x^2 + \frac{8}{x} = 0 \Rightarrow x^3 + 8 = 0 \Rightarrow x^3 = -8 \Rightarrow x = -2$.For $x = -2$,$y = \frac{2}{-2} = -1$. So,the point of intersection is $(-2, -1)$.Differentiating $(i)$ with respect to $x$:$x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.At $(-2, -1)$,$m_1 = -(\frac{-1}{-2}) = -\frac{1}{2}$.Differentiating $(ii)$ with respect to $x$:$2x + 4 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{2x}{4} = -\frac{x}{2}$.At $(-2, -1)$,$m_2 = -(\frac{-2}{2}) = 1$.The angle $	heta$ between the curves is given by:$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{-\frac{1}{2} - 1}{1 + (-\frac{1}{2})(1)} ight| = \left| \frac{-\frac{3}{2}}{1 - \frac{1}{2}} ight| = \left| \frac{-\frac{3}{2}}{\frac{1}{2}} ight| = |-3| = 3$.Therefore,$	an 	heta = 3$.

If $\theta$ is the angle between the curves $xy=2$ and $x^2+4y=0$,then $\tan \theta$ is equal to:

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