Question

Find the angle of intersection of the following curve x² + y² = 2x and y² = x ?

Answer 1

\[\text {  Given curves are},\]

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

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

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

\[ x^2 + x = 2x\]

\[ \Rightarrow x^2 - x = 0\]

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

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

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

\[y = 0 \text { or} y=\pm1 \]

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

\[\text { Differentiating (1) w.r.t.x,we get},\]

\[2x + 2y\frac{dy}{dx} = 2\]

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

\[\text { Differentiating (2) w.r.t. x,we get },\]

\[2y \frac{dy}{dx} = 1\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2y} . . . \left( 4 \right)\]

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

\[\text { From } \left( 3 \right) \text { we get, m_1 is undefined }. \]

\[ \therefore \text { We can not find } \theta\]

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

\[\text { From } \left( 3 \right) \text { we get,} m_1 = 0\]

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

\[\text { Now }, \]

\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{0 - \frac{1}{2}}{1 + 0} \right| = \frac{1}{2}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{1}{2} \right)\]

\[\text {Case } -3:\text { Let } \left( x, y \right) =\left( 1, - 1 \right)\]

\[\text { From }  \left( 3 \right)\text { we get }, m_1 = 0\]

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

\[\text { Now, } \]

\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{0 + \frac{1}{2}}{1} \right| = \frac{1}{2}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{1}{2} \right)\]