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

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(C) Given curves are $y^2=2x$ and $x^2+y^2=8$.Substituting $y^2=2x$ into the second equation: $x^2+2x-8=0$.Factoring gives $(x+4)(x-2)=0$. Since $x$ m

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$	an^{-1}(3)$

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(C) Given curves are $y^2=2x$ and $x^2+y^2=8$.Substituting $y^2=2x$ into the second equation: $x^2+2x-8=0$.Factoring gives $(x+4)(x-2)=0$. Since $x$ must be non-negative for $y^2=2x$,we have $x=2$.For $x=2$,$y^2=4$,so $y=2$ (taking the positive intersection point).Differentiating $y^2=2x$ gives $2y \frac{dy}{dx} = 2$,so $m_1 = \frac{dy}{dx} = \frac{1}{y} = \frac{1}{2}$ at $(2, 2)$.Differentiating $x^2+y^2=8$ gives $2x + 2y \frac{dy}{dx} = 0$,so $m_2 = \frac{dy}{dx} = -\frac{x}{y} = -\frac{2}{2} = -1$ at $(2, 2)$.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{1/2 - (-1)}{1 + (1/2)(-1)} ight| = \left| \frac{3/2}{1/2} ight| = 3$.Therefore,$	heta = 	an^{-1}(3)$.

The angle between the curves $y^2=2x$ and $x^2+y^2=8$ is

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