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Find the angle of intersection of the curve \( y^2=x \: and \: x^2=y\)

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  • $ \tan\theta=\bigg|\large\frac{m_1-m_2}{1+m_1m_2}\bigg|$
Step 1:
Let the curve be $y^2=x$
Differentiating with respect to x we get,
$\Rightarrow \large\frac{dy}{dx}=\frac{1}{2y}$
Let this be $m_1$
$\therefore m_1=\large\frac{1}{2y}$
Step 2:
Let the other curve be $x^2=y$
Differentiating with respect to $x$ we get,
Let this be $m_2$
$\therefore m_2=2x$
Step 3:
The points of intersection of the curves are
$\Rightarrow x(x^3-1)=0$
Hence $x=0$ or $x=1$
When $x=0,y=0$
When $x=1,y=1$
Hence the points of intersection are $(0,0)$ and $(1,1)$
Step 4:
Now $m_1$ at $(1,1)$ is
$\Rightarrow \large\frac{1}{1}$$=1$
Now $m_2$ at $(1,1)$ is
$\Rightarrow 2(1)=2$
Step 5:
$\therefore \tan\theta=\bigg|\large\frac{m_1-m_2}{1+m_1m_2}\bigg|$
answered Oct 3, 2013 by sreemathi.v
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