Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • $ \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
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App