What is the angle of intersection of the curv | vedclass

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(C) To find the intersection points,set $4 - x^2 = x^2$,which gives $2x^2 = 4$,so $x^2 = 2$,$x = \pm \sqrt{2}$.For $x = \pm \sqrt{2}$,$y = 2$. Thus,th

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$	an^{-1}(4\sqrt{2} / 7)$

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(C) To find the intersection points,set $4 - x^2 = x^2$,which gives $2x^2 = 4$,so $x^2 = 2$,$x = \pm \sqrt{2}$.For $x = \pm \sqrt{2}$,$y = 2$. Thus,the points of intersection are $(\sqrt{2}, 2)$ and $(-\sqrt{2}, 2)$.For the curve $y = 4 - x^2$,the slope $m_1 = \frac{dy}{dx} = -2x$.For the curve $y = x^2$,the slope $m_2 = \frac{dy}{dx} = 2x$.At the point $(\sqrt{2}, 2)$,$m_1 = -2\sqrt{2}$ and $m_2 = 2\sqrt{2}$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.$	an 	heta = \left| \frac{-2\sqrt{2} - 2\sqrt{2}}{1 + (-2\sqrt{2})(2\sqrt{2})} ight| = \left| \frac{-4\sqrt{2}}{1 - 8} ight| = \left| \frac{-4\sqrt{2}}{-7} ight| = \frac{4\sqrt{2}}{7}$.Therefore,$	heta = 	an^{-1}\left( \frac{4\sqrt{2}}{7} ight)$.

What is the angle of intersection of the curves $y = 4 - x^2$ and $y = x^2$?

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