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

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(D) The equations of the curves are $xy=2$ $\dots(i)$ and $x^2+4y=0$ $\dots(ii)$.To find the point of intersection,substitute $y = -x^2/4$ from $(ii)$

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$3$

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(D) The equations of the curves are $xy=2$ $\dots(i)$ and $x^2+4y=0$ $\dots(ii)$.To find the point of intersection,substitute $y = -x^2/4$ from $(ii)$ into $(i)$:$x(-x^2/4) = 2 \Rightarrow -x^3 = 8 \Rightarrow x = -2$.Substituting $x = -2$ into $(ii)$,we get $4 + 4y = 0 \Rightarrow y = -1$.So,the point of intersection is $(-2, -1)$.For curve $(i)$,$y = 2/x$,so $dy/dx = -2/x^2$. At $x = -2$,$m_1 = -2/(-2)^2 = -2/4 = -1/2$.For curve $(ii)$,$x^2 + 4y = 0$,so $2x + 4(dy/dx) = 0 \Rightarrow dy/dx = -x/2$. At $x = -2$,$m_2 = -(-2)/2 = 1$.The angle $	heta$ between the curves is given by $	an 	heta = |(m_1 - m_2) / (1 + m_1 m_2)|$.$	an 	heta = |(-1/2 - 1) / (1 + (-1/2)(1))| = |(-3/2) / (1/2)| = |-3| = 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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