If theta is the angle between the curves x^2- | vedclass

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(B) Given curves are $x^2-y^2=4$ and $y^2=3x$. Substituting $y^2=3x$ into the first equation: $x^2-3x-4=0$. $(x-4)(x+1)=0$. Since $y^2=3x$,$x$ must be

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$\frac{5}{6 \sqrt{3}}$

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(B) Given curves are $x^2-y^2=4$ and $y^2=3x$. Substituting $y^2=3x$ into the first equation: $x^2-3x-4=0$. $(x-4)(x+1)=0$. Since $y^2=3x$,$x$ must be positive,so $x=4$. Then $y^2=12$,so $y=\pm 2\sqrt{3}$. Let the point of intersection be $(4, 2\sqrt{3})$. For $x^2-y^2=4$,differentiating w.r.t $x$: $2x-2y \frac{dy}{dx}=0 \implies \frac{dy}{dx}=\frac{x}{y}$. At $(4, 2\sqrt{3})$,$m_1 = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$. For $y^2=3x$,differentiating w.r.t $x$: $2y \frac{dy}{dx}=3 \implies \frac{dy}{dx}=\frac{3}{2y}$. At $(4, 2\sqrt{3})$,$m_2 = \frac{3}{2(2\sqrt{3})} = \frac{3}{4\sqrt{3}} = \frac{\sqrt{3}}{4}$. $	an 	heta = \left| \frac{m_1-m_2}{1+m_1 m_2} ight| = \left| \frac{\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{4}}{1+(\frac{2}{\sqrt{3}})(\frac{\sqrt{3}}{4})} ight| = \left| \frac{\frac{8-3}{4\sqrt{3}}}{1+\frac{1}{2}} ight| = \left| \frac{5}{4\sqrt{3}} 	imes \frac{2}{3} ight| = \frac{5}{6\sqrt{3}}$.

If $\theta$ is the angle between the curves $x^2-y^2=4$ and $y^2=3x$,then $\tan \theta=$

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