Question

Find the angle of intersection of the following curve y² = x and x² = y  ?

Answer 1

\[\text { Given curves are },\]

\[ y^2 = x . . . \left( 1 \right)\]

\[ x^2 = y . . . \left( 2 \right)\]

\[\text { From these two equations, we get }\]

\[ \left( x^2 \right)^2 = x\]

\[ \Rightarrow x^4 - x = 0\]

\[ \Rightarrow x \left( x^3 - 1 \right) = 0\]

\[ \Rightarrow x = 0 orx= 1\]

\[\text { Substituting the values of x in } \left( 2 \right) \text { we get }, \]

\[y = 0 ory = 1 \]

\[ \therefore \left( x, y \right)=\left( 0, 0 \right) or \left( 1, 1 \right)\]

\[\text { Differenntiating (1) w.r.t.x},\]

\[2y \frac{dy}{dx} = 1\] 

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2y} . . . \left( 3 \right)\]

\[\text { Differenntiating(2) w.r.t.x },\]

\[2x = \frac{dy}{dx} . . . \left( 4 \right)\]

\[Case -1: \left( x, y \right)=\left( 0, 0 \right)\]

\[\text { The tangent to curve is parallel to x - axis } . \]

\[\text { Hence, the angle between the tangents to two curve at } \left( 0, 0 \right) \text { is a right angle} . \]

\[ \therefore \theta = \frac{\pi}{2}\]

\[\text { Case }-2: \left( x, y \right)=\left( 1, 1 \right)\]

\[\text { From } \left( 3 \right) \text { we have }, m_1 = \frac{1}{2}\]

\[\text { From } \left( 4 \right) \text { we have }, m_2 = 2 \left( 1 \right) = 2\]

\[\text { Now,} \]

\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{1}{2} - 2}{1 + \frac{1}{2} \times 2} \right| = \frac{3}{4}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{3}{4} \right)\]