1708Yes Narayan You are Right.
\dpi{120} \hspace{-16}(1)::\; $Find acute Angle b/w 2 -Curves $\mathbf{C_{1}:x^2+y^2=8}$ and $\mathbf{C_{2}:x^2+y^2=4x}$.\\\\ (2)::\;Find Least distance b/w the curves $\mathbf{xy=9}$ and $\mathbf{x^2+y^2=1}$.
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5 Answers
1x^2+y^2=8\rightarrow 2x+2y.\frac{dy}{dx} =0\rightarrow x+y\left ( \frac{dy}{dx} \right ) =0\rightarrow \frac{dy}{dx} =\frac{-x}{y}=m_{1}
x^2+y^2=4x\rightarrow 2x+2y\left ( \frac{dy}{dx} \right )=4\rightarrow x+y\left ( \frac{dy}{dx} \right )=2\rightarrow \frac{dy}{dx} =\frac{2-x}{y}=m_{2}
\textup{solving for intersection we get}\, \, 4x=8\Rightarrow x=2
\textup{when} \, \, x=2 , y=\pm 2
\textup{when} \, \, x=2 , y= 2 \rightarrow m _{1} =-1,\, \, m_{2}=0
tan\theta =\left | \frac{m_{1}-m_{2}}{1+m_{1}.m_{2}} \right |
tan\theta =\left | \frac{m_{1}-m_{2}}{1+m_{1}.m_{2}} \right |=\left | \frac{-1-0}{1+0} \right |= 1 \Rightarrow \theta =\frac{\pi }{4}
\textup{when} \, \, x=2 , y= -2 \rightarrow m _{1} =1,\, \, m_{2}=0
tan\theta =\left | \frac{m_{1}-m_{2}}{1+m_{1}.m_{2}} \right |=\left | \frac{1-0}{1+0} \right |=1\Rightarrow \theta =\frac{\pi }{4}
\textup{ Acute Angle between the curves = }\, \frac{\pi }{4}=45^{\circ}
2622) it is easily understood that min. distance will be when x=y (from graph)
therefore min. distance= 3√2-1
