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

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(B) Given curves are $x^2-y^2=4$ ...$(i)$ and $x^2+y^2=4\sqrt{2}$ ...(ii).Adding $(i)$ and (ii), we get $2x^2 = 4(1+\sqrt{2})$, so $x^2 = 2(1+\sqrt{2}

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$\frac{\pi}{4}$

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(B) Given curves are $x^2-y^2=4$ ...$(i)$ and $x^2+y^2=4\sqrt{2}$ ...(ii).Adding $(i)$ and (ii), we get $2x^2 = 4(1+\sqrt{2})$, so $x^2 = 2(1+\sqrt{2})$.Subtracting $(i)$ from (ii), we get $2y^2 = 4(\sqrt{2}-1)$, so $y^2 = 2(\sqrt{2}-1)$.Differentiating $(i)$ with respect to $x$: $2x - 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y} = m_1$.Differentiating (ii) with respect to $x$: $2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y} = m_2$.The angle $	heta$ between the curves is given by $	an 	heta = \left|\frac{m_1-m_2}{1+m_1m_2}ight|$.Substituting the values: $	an 	heta = \left|\frac{\frac{x}{y} - (-\frac{x}{y})}{1 + (\frac{x}{y})(-\frac{x}{y})}ight| = \left|\frac{2x/y}{1 - x^2/y^2}ight| = \left|\frac{2xy}{y^2-x^2}ight|$.From $(i)$ and (ii), $x^2-y^2=4$ and $x^2+y^2=4\sqrt{2}$. Thus $y^2-x^2 = -4$.Also, $x^2y^2 = 4(\sqrt{2}+1) 	imes 2(\sqrt{2}-1) = 8(2-1) = 8$, so $xy = \sqrt{8} = 2\sqrt{2}$.Therefore, $	an 	heta = \left|\frac{2(2\sqrt{2})}{-4}ight| = \left|-\sqrt{2}ight|$ is incorrect in the original prompt logic; let's re-evaluate: $	an 	heta = \left|\frac{2xy}{y^2-x^2}ight| = \left|\frac{2(2\sqrt{2})}{-4}ight| = \sqrt{2}$. Wait, checking the intersection: $x^2 = 2+2\sqrt{2}$, $y^2 = 2\sqrt{2}-2$. $y^2-x^2 = -4$. $x^2y^2 = 4(2-1) = 4$, so $xy=2$. $	an 	heta = |4/-4| = 1$.Thus, $	heta = \frac{\pi}{4}$.

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

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