The angle of intersection between the curves | vedclass

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

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$\frac{\pi}{4}$

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(B) Given curves are $y^2+x^2=a^2 \sqrt{2}$ $(i)$ and $x^2-y^2=a^2$ $(ii)$.Adding $(i)$ and $(ii)$,we get $2x^2 = a^2(\sqrt{2}+1) \Rightarrow x^2 = \frac{a^2(\sqrt{2}+1)}{2}$.Subtracting $(ii)$ from $(i)$,we get $2y^2 = a^2(\sqrt{2}-1) \Rightarrow y^2 = \frac{a^2(\sqrt{2}-1)}{2}$.For curve $(i)$,differentiating with respect to $x$: $2x + 2y \frac{dy}{dx} = 0 \Rightarrow m_1 = \frac{dy}{dx} = -\frac{x}{y}$.For curve $(ii)$,differentiating with respect to $x$: $2x - 2y \frac{dy}{dx} = 0 \Rightarrow m_2 = \frac{dy}{dx} = \frac{x}{y}$.The angle of intersection $	heta$ is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.Substituting $m_1 = -\frac{x}{y}$ and $m_2 = \frac{x}{y}$:$	an 	heta = |\frac{-\frac{x}{y} - \frac{x}{y}}{1 + (-\frac{x}{y})(\frac{x}{y})}| = |\frac{-2x/y}{1 - x^2/y^2}| = |\frac{-2xy}{y^2 - x^2}|$.Since $x^2 = \frac{a^2(\sqrt{2}+1)}{2}$ and $y^2 = \frac{a^2(\sqrt{2}-1)}{2}$,then $y^2 - x^2 = -a^2$.Also $x^2 y^2 = \frac{a^4(2-1)}{4} = \frac{a^4}{4} \Rightarrow xy = \frac{a^2}{2}$.$	an 	heta = |\frac{-2(a^2/2)}{-a^2}| = |\frac{-a^2}{-a^2}| = 1$.Therefore,$	heta = \frac{\pi}{4}$.

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

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