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