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

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Aicharya · Accepted Answer

Given, curves are : $x^2+y^2=4 x$ ...(i) and$x^2+y^2=8$ ...(ii) $\because \frac{d y}{d x}$ is the slope of tangent drawn to curve Slope of the tangent drawn to curve (i) is $m_1=\frac{d y}{d x} \quad$ [from curve (i)] On differentiating Eq. (i) w.r.t. $x$ $\frac{d}{d x}\left(x^2+y^2ight)=\frac{d}{d x}(4 x)$ $\Rightarrow \quad 2 x+2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{4-2 x}{2 y}$ $\left.	herefore \quad m_1ight|_{a t(2,2)}=\frac{4-2 	imes 2}{2 	imes 2}=0$ On differentiating Eq. (ii) w.r.t. $x$, $\frac{d}{d x}\left(x^2+y^2ight)=\frac{d}{d x}(8)$ $\begin{array}{ll}\Rightarrow & 2 x+2 y \frac{d y}{d x}=0 \ \Rightarrow & \frac{d y}{d x}=\frac{-2 x}{2 y}=\frac{-x}{y}\end{array}$ $\left.m_2ight|_{(2,2)}=\frac{-2}{2}=-1$ $	herefore \quad 	an 	heta=\left|\frac{m_2-m_1}{1+m_1 \cdot m_2}ight|$ $=\left|\frac{-1-0}{1+0}ight|=1$ $\Rightarrow \quad 	heta=	an ^{-1}(\mathrm{l})=\frac{\pi}{4}$ $\because \quad 	heta=\sin ^{-1}(a) \quad$ [Given] $\Rightarrow \quad \sin ^{-1}(a)=\frac{\pi}{4}$ $\Rightarrow \quad a=\sin \frac{\pi}{4} \Rightarrow a=\frac{1}{\sqrt{2}}$

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

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