Match the following pairof curveswith their angleof intersections

a)  $x^2+y^2=2{\pi }^2$

$y={\sin }^{-1}[x^2+\frac{1}{2}]+{\cos }^{-1}[x^2-\frac{1}{2}]\Rightarrow y=k$

$\therefore$does not exist

b) Range of  $|\sin x|+|\cos x|=[1,\sqrt{2}]$

$\Rightarrow \left[\right|\sin x|+|\cos x\left|\right]=1\Rightarrow y=1$

$\Rightarrow y^2=2x\Rightarrow x=\frac{1}{2}$

Point of intersection  $p=(\frac{1}{2},1)$

$\Rightarrow \frac{dy}{dx}=\frac{-(-2)}{2y}=\frac{1}{y}\Rightarrow m=1\Rightarrow \mathrm{Tan}\theta =1$

$\Rightarrow \theta =\frac{\pi }{4}\text{(or)}\frac{3\pi }{4}$

c)  $x^2=4ay,y=\frac{8a^3}{x^2+4a^2}$

$\Rightarrow x^2-4ay=0,x^{2}y+4a^{2}y-8a^3=0$   point of  intersection=  $(\pm 2a,a)$

$y=\frac{x^2}{4a}\Rightarrow m_1=1 y=\frac{8a^3}{x^2+4a^2}\Rightarrow m_2=-1/2$;

$\mathrm{Tan}\theta =\left|\frac{1+1/2}{1-1/2}\right|=3\Rightarrow \theta ={\mathrm{Tan}}^{-1}\left(3\right){\cot }^{-1}(1/3)$

 d) From the two curves  $\frac{2x}{\pi }=\sin x\Rightarrow x=\frac{\pi }{2}\Rightarrow y=1$

$\Rightarrow (\frac{\pi }{2},1)y^2=\frac{2x}{\pi }\Rightarrow m_1=\frac{1}{\pi }$

$y=\sin x\Rightarrow \frac{dy}{dx}=\cos x\Rightarrow m_2=0$

$\mathrm{Tan}\theta =\left|\frac{\frac{1}{\pi }-0}{1+0\left(\frac{1}{\pi }\right)}\right|=\frac{1}{\pi }\Rightarrow \theta ={\mathrm{Tan}}^{-1}\left(\frac{1}{\pi }\right)={\cot }^{-1}\left(\pi \right)$