If theta is an angle between the curves x^2+4 | vedclass

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(D) To find the point of intersection of the curves $x^2+4y=0$ and $xy=2$:Substitute $y = \frac{2}{x}$ into $x^2+4y=0$:$x^2 + 4(\frac{2}{x}) = 0$$x^2

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$3$

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(D) To find the point of intersection of the curves $x^2+4y=0$ and $xy=2$:Substitute $y = \frac{2}{x}$ into $x^2+4y=0$:$x^2 + 4(\frac{2}{x}) = 0$$x^2 + \frac{8}{x} = 0$$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 $(-2, -1)$.Slope of the first curve $x^2+4y=0$:Differentiating with respect to $x$: $2x + 4\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{2}$.At $x = -2$,$m_1 = -\frac{-2}{2} = 1$.Slope of the second curve $xy=2$:Differentiating with respect to $x$: $y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.At $(-2, -1)$,$m_2 = -\frac{-1}{-2} = -\frac{1}{2}$.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{1 - (-1/2)}{1 + (1)(-1/2)} ight| = \left| \frac{3/2}{1/2} ight| = 3$.

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

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