If sin ^{-1}(a) is the acute angle between th | vedclass

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(B) Given curves are: $x^2+y^2=4x$  $(i)$ and $x^2+y^2=8$ $(ii)$.To find the angle between the curves at $(2,2)$,we find the slopes of the tangents at

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$\frac{1}{\sqrt{2}}$

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(B) Given curves are: $x^2+y^2=4x$  $(i)$ and $x^2+y^2=8$ $(ii)$.To find the angle between the curves at $(2,2)$,we find the slopes of the tangents at this point.For curve $(i)$,differentiating with respect to $x$: $2x + 2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2-x}{y}$.At $(2,2)$,$m_1 = \frac{2-2}{2} = 0$.For curve $(ii)$,differentiating with respect to $x$: $2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$.At $(2,2)$,$m_2 = -\frac{2}{2} = -1$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight|$.$	an 	heta = \left| \frac{-1 - 0}{1 + 0 	imes (-1)} ight| = |-1| = 1$.Thus,$	heta = 	an^{-1}(1) = \frac{\pi}{4}$.Given $	heta = \sin^{-1}(a)$,so $\sin^{-1}(a) = \frac{\pi}{4}$.Therefore,$a = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

If $\sin ^{-1}(a)$ is the acute angle between the curves $x^2+y^2=4x$ and $x^2+y^2=8$ at the point $(2,2)$,then $a$ is equal to

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