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Question
Find the angle of intersection of the following curve x2 + y2 = 2x and y2 = x ?
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Solution
\[\text { Given curves are},\]
\[ x^2 + y^2 = 2x . . . \left( 1 \right)\]
\[ y^2 = x . . . \left( 2 \right)\]
\[\text { From these two equations we get }\]
\[ x^2 + x = 2x\]
\[ \Rightarrow x^2 - x = 0\]
\[ \Rightarrow x \left( x - 1 \right) = 0\]
\[ \Rightarrow x = 0 orx = 1\]
\[\text { Substituting the values of x in } \left( 2 \right) \text { we get }, \]
\[y = 0 \text { or} y=\pm1 \]
\[\therefore\left( x, y \right) =\left( 0, 0 \right),\left( 1, 1 \right),\left( 1, - 1 \right)\]
\[\text { Differentiating (1) w.r.t.x,we get},\]
\[2x + 2y\frac{dy}{dx} = 2\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1 - x}{y} . . . \left( 3 \right)\]
\[\text { Differentiating (2) w.r.t. x,we get },\]
\[2y \frac{dy}{dx} = 1\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2y} . . . \left( 4 \right)\]
\[\text { Case }-1: \left( x, y \right) =\left( 0, 0 \right)\]
\[\text { From } \left( 3 \right) \text { we get, m_1 is undefined }. \]
\[ \therefore \text { We can not find } \theta\]
\[\text { Case } -2:Let \left( x, y \right) =\left( 1, 1 \right)\]
\[\text { From } \left( 3 \right) \text { we get,} m_1 = 0\]
\[\text { From } \left( 4 \right) \text { we get,} m_2 = \frac{1}{2}\]
\[\text { Now }, \]
\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{0 - \frac{1}{2}}{1 + 0} \right| = \frac{1}{2}\]
\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{1}{2} \right)\]
\[\text {Case } -3:\text { Let } \left( x, y \right) =\left( 1, - 1 \right)\]
\[\text { From } \left( 3 \right)\text { we get }, m_1 = 0\]
\[\text { From } \left( 4 \right) \text { we get,} m_2 = \frac{- 1}{2}\]
\[\text { Now, } \]
\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{0 + \frac{1}{2}}{1} \right| = \frac{1}{2}\]
\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{1}{2} \right)\]
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