An angle between the curves x^2-y^2=4 and x^2 | vedclass

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(B) Given curves are $x^2-y^2=4$ and $x^2+y^2=4 \sqrt{2}$.Let $(x_0, y_0)$ be the point of intersection.The angle between the curves is the angle betw

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$\frac{\pi}{4}$

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(B) Given curves are $x^2-y^2=4$ and $x^2+y^2=4 \sqrt{2}$.Let $(x_0, y_0)$ be the point of intersection.The angle between the curves is the angle between their tangents at the point of intersection.For the first curve $x^2-y^2=4$,differentiating with respect to $x$ gives $2x - 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = \frac{x}{y}$. At $(x_0, y_0)$,$m_1 = \frac{x_0}{y_0}$.For the second curve $x^2+y^2=4 \sqrt{2}$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{x}{y}$. At $(x_0, y_0)$,$m_2 = -\frac{x_0}{y_0}$.The angle $	heta$ between the tangents is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the slopes: $	an 	heta = \left| \frac{\frac{x_0}{y_0} - (-\frac{x_0}{y_0})}{1 + (\frac{x_0}{y_0})(-\frac{x_0}{y_0})} ight| = \left| \frac{2 \frac{x_0}{y_0}}{1 - \frac{x_0^2}{y_0^2}} ight| = \left| \frac{2 x_0 y_0}{y_0^2 - x_0^2} ight|$.Since $x_0^2 - y_0^2 = 4$,we have $y_0^2 - x_0^2 = -4$.Thus,$	an 	heta = \left| \frac{2 x_0 y_0}{-4} ight| = \frac{|x_0 y_0|}{2}$.Solving the system $x_0^2 - y_0^2 = 4$ and $x_0^2 + y_0^2 = 4 \sqrt{2}$:Adding the equations: $2x_0^2 = 4(1 + \sqrt{2}) \Rightarrow x_0^2 = 2(1 + \sqrt{2})$.Subtracting the equations: $2y_0^2 = 4(\sqrt{2} - 1) \Rightarrow y_0^2 = 2(\sqrt{2} - 1)$.Then $x_0^2 y_0^2 = 4(1 + \sqrt{2})(\sqrt{2} - 1) = 4(2 - 1) = 4$.So,$|x_0 y_0| = 2$.Substituting this into the expression for $	an 	heta$: $	an 	heta = \frac{2}{2} = 1$.Therefore,$	heta = 	an^{-1}(1) = \frac{\pi}{4}$.

An angle between the curves $x^2-y^2=4$ and $x^2+y^2=4 \sqrt{2}$ is

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