**Angle of Intersection of Two Curves **

The angle of intersection of two curves is the angle subtended between the tangents at their point of intersection.

Let C₁ and C₂ be two curves having equations y = f(x) and y = g(x), respectively.

Let PT₁ and PT₂ be tangents to the curves C₁ and C₂ at their point of intersection.

Let θ be the angle between the two tangents PT₁ and PT₂ and θ₁ and θ₂ are the angles made by tangents with the positive direction of x-axis in anti-clockwise sense.

Then, \({{m}_{1}}=\tan {{\theta }_{1}}=\left( \frac{dy}{dx} \right){{C}_{1}}\) and \({{m}_{2}}=\tan {{\theta }_{2}}=\left( \frac{dy}{dx} \right){{C}_{2}}\).

From the figure it follows, θ = θ₂ – θ₁

\(\tan \theta =\tan \left( {{\theta }_{2}}-{{\theta }_{1}} \right)=\frac{\tan {{\theta }_{2}}-\tan {{\theta }_{1}}}{1+\tan {{\theta }_{2}}\tan {{\theta }_{1}}}\),

\(\tan \theta =\left| \frac{{{\left( \frac{dy}{dx} \right)}_{{{C}_{1}}}}-{{\left( \frac{dy}{dx} \right)}_{{{C}_{2}}}}}{1+{{\left( \frac{dy}{dx} \right)}_{{{C}_{1}}}}{{\left( \frac{dy}{dx} \right)}_{{{C}_{2}}}}} \right|=\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\).

Angle of intersection of these curves is defined as acute angle between the tangents.

**Example: **The acute angle between the curves y = |x² – 1| and y = |x² – 3| at their points of intersections is.

**Solution: **The points of intersection are (±√2, 1)

Since, the curves are symmetrical about y-axis, the angle of intersection at (-√2, 1)

= The angle of intersection at (√2, 1).

At (√2, 1); m₁ = 2x = 2√2; m₂ = – 2x = -2√2.

\(\tan \theta =\left| \frac{4\sqrt{2}}{1-8} \right|=\frac{4\sqrt{2}}{7}\),

\(\theta ={{\tan }^{-1}}\frac{4\sqrt{2}}{7}\).