Browse Questions

# Find the angle of intersection of the curve $y^2=x \: and \: x^2=y$

Toolbox:
• $\tan\theta=\bigg|\large\frac{m_1-m_2}{1+m_1m_2}\bigg|$
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:
$\therefore \tan\theta=\bigg|\large\frac{m_1-m_2}{1+m_1m_2}\bigg|$
$\qquad\quad=\bigg|\large\frac{1-2}{1+2}\bigg|$
$\tan\theta=\bigg|\large\frac{-1}{3}\bigg|$
$\theta=\tan^{-1}\big(\large\frac{1}{3}\big)$