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Question
Find the angle of intersection of the following curve x2 + 4y2 = 8 and x2 − 2y2 = 2 ?
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Solution
\[\text { Given curves are },\]
\[ x^2 + 4 y^2 = 8 . . . \left( 1 \right)\]
\[ x^2 - 2 y^2 = 2 . . . \left( 2 \right)\]
\[\text { From (1) and (2) we get }\]
\[6 y^2 = 6\]
\[ \Rightarrow y = 1 or y_1 = - 1\]
\[\text { Substituting the values of y in eq.} \left( 1 \right)\]
\[x = 2, - 2 orx = 2, - 2 \]
\[\text { So},\left( x, y \right)=\left( 2, 1 \right),\left( 2, - 1 \right),\left( - 2, 1 \right),\left( - 2, - 1 \right)\]
\[\text { Differentiating (1) w.r.t.x },\]
\[2x + 8y \frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- x}{4y} . . . \left( 3 \right)\]
\[\text { Differentiating (2) w.r.t.x },\]
\[2x - 4y \frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x}{2y} . . . \left( 4 \right) \]
\[ \text { Case } -1: \left( x, y \right)=\left( 2, 1 \right)\]
\[\text { From} \left( 3 \right), \text { we get, } m_1 = \frac{- 1}{2}\]
\[\text { From} \left( 4 \right), \text { we get,} m_2 = 1\]
\[\text { We have,} \]
\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}} \right| = 3\]
\[ \Rightarrow \theta = \tan^{- 1} \left( 3 \right)\]
\[\text { Case } -2: \left( x, y \right)=\left( 2, - 1 \right)\]
\[\text { From } \left( 3 \right),\text { we get, } m_1 = \frac{1}{2}\]
\[\text { From } \left( 4 \right), \text { we get,} m_2 = - 1\]
\[\text { We have,} \]
\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{1}{2} + 1}{1 - \frac{1}{2}} \right| = 3\]
\[ \Rightarrow \theta = \tan^{- 1} \left( 3 \right)\]
\[\text { Case } -3: \left( x, y \right)=\left( - 2, 1 \right)\]
\[\text { From } \left( 3 \right),\text { we get, } m_1 = \frac{1}{2}\]
\[\text { From } \left( 4 \right),\text { we get,} m_2 = - 1\]
\[\text { We have}, \]
\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{1}{2} + 1}{1 - \frac{1}{2}} \right| = 3\]
\[ \Rightarrow \theta = \tan^{- 1} \left( 3 \right)\]
\[\text { Case } -4: \left( x, y \right)=\left( - 2, - 1 \right)\]
\[\text { From } \left( 3 \right), \text { we get,} m_1 = \frac{- 1}{2}\]
\[\text { From} \left( 4 \right), \text { we get,} m_2 = 1\]
\[\text { We have,} \]
\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}} \right| = 3\]
\[ \Rightarrow \theta = \tan^{- 1} \left( 3 \right)\]
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