An angle between the curves x^2=3y and x^2+y^ | vedclass

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(A) Given curves are:$x^2 = 3y \quad ...(i)$$x^2 + y^2 = 4 \quad ...(ii)$Substituting $x^2 = 3y$ into equation $(ii)$,we get:$3y + y^2 = 4$$y^2 + 3y -

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$	an ^{-1} \frac{5}{\sqrt{3}}$

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(A) Given curves are:$x^2 = 3y \quad ...(i)$$x^2 + y^2 = 4 \quad ...(ii)$Substituting $x^2 = 3y$ into equation $(ii)$,we get:$3y + y^2 = 4$$y^2 + 3y - 4 = 0$$(y + 4)(y - 1) = 0$Since $x^2 = 3y$,$y$ must be non-negative,so $y = 1$.Substituting $y = 1$ into $x^2 = 3y$,we get $x^2 = 3$,so $x = \pm \sqrt{3}$.Thus,the points of intersection are $(\sqrt{3}, 1)$ and $(-\sqrt{3}, 1)$.Now,differentiating the curves with respect to $x$:For $(i)$,$2x = 3 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{2x}{3}$.For $(ii)$,$2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$.At the point $(\sqrt{3}, 1)$:$m_1 = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$$m_2 = -\frac{\sqrt{3}}{1} = -\sqrt{3}$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{2}{\sqrt{3}} - (-\sqrt{3})}{1 + (\frac{2}{\sqrt{3}})(-\sqrt{3})} ight|$$	an 	heta = \left| \frac{\frac{2 + 3}{\sqrt{3}}}{1 - 2} ight| = \left| \frac{5/\sqrt{3}}{-1} ight| = \frac{5}{\sqrt{3}}$Therefore,$	heta = 	an^{-1} \left( \frac{5}{\sqrt{3}} ight)$.

An angle between the curves $x^2=3y$ and $x^2+y^2=4$ is

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