If theta is the acute angle between the curve | vedclass

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(A) 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 $x$ must be non-negative

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

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(A) 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 $x$ must be non-negative for $y^2=3x$, we have $x=1$.For $x=1$, $y^2=3$, so $y=\sqrt{3}$ (considering the point of intersection in the first quadrant).Now, differentiate both curves with respect to $x$:For $x^2+y^2=4$, $2x+2y\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$. At $(1, \sqrt{3})$, $m_1 = -\frac{1}{\sqrt{3}}$.For $y^2=3x$, $2y\frac{dy}{dx}=3 \Rightarrow \frac{dy}{dx} = \frac{3}{2y}$. At $(1, \sqrt{3})$, $m_2 = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.The angle $	heta$ between the curves is given by $	an 	heta = \left|\frac{m_1-m_2}{1+m_1m_2}ight|$.$	an 	heta = \left|\frac{-\frac{1}{\sqrt{3}} - \frac{\sqrt{3}}{2}}{1 + (-\frac{1}{\sqrt{3}})(\frac{\sqrt{3}}{2})}ight| = \left|\frac{-\frac{2+3}{2\sqrt{3}}}{1-\frac{1}{2}}ight| = \left|\frac{-\frac{5}{2\sqrt{3}}}{\frac{1}{2}}ight| = \frac{5}{\sqrt{3}}$.

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

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