# Find the angle of intersection of the curves

Question:

Find the angle of intersection of the curves = 4 – x2 and x2.

Solution:

The given curves are = 4 – x2 …. (i) and x2 …… (ii)

And, we know that the angle of intersection of two curves is equal to the angle between the tangents drawn to the curves at their point of intersection.

Now, differentiating equations (i) and (ii) w.r.t x, we have

dy/dx = -2x ⇒ m1 = -2x

m1 is the slope of the tangent to the curve (i).

and dy/dx = 2x ⇒ m2 = 2x

m2 is the slope of the tangent to the curve (ii).

So, m1 = -2x and m2 = 2x

On solving equation (i) and (ii), we get

4 – x2 = x2 ⇒ 2x2 = 4 ⇒ x2 = 2 ⇒ x = ±√2

So, m1 = -2x = -2√2 and m2 = 2x = 2√2

Let θ be the angle of intersection of two curves

So, $\quad \tan \theta=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|$

$=\left|\frac{2 \sqrt{2}+2 \sqrt{2}}{1-(2 \sqrt{2})(2 \sqrt{2})}\right|=\left|\frac{4 \sqrt{2}}{1-8}\right|=\left|\frac{4 \sqrt{2}}{-7}\right|=\frac{4 \sqrt{2}}{7}$

$\therefore \theta=\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)$

Therefore, the required angle is $\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)$.