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Question
Find the angle of intersection of the following curve y = 4 − x2 and y = x2 ?
Solution
\[\text { Given curves are},\]
\[y = 4 - x^2 . . . . . \left( 1 \right)\]
\[y = x^2 . . . . . \left( 2 \right)\]
\[\text { From ( 1)and (2), we get }\]
\[4 - x^2 = x^2 \]
\[ \Rightarrow 2 x^2 = 4\]
\[ \Rightarrow x^2 = 2\]
\[ \Rightarrow x = \pm \sqrt{2}\]
\[\text { Substituting the values of x in (2), we get }, \]
\[ \Rightarrow y = 2\]
\[ \Rightarrow \left( x, y \right)=\left( \sqrt{2},2 \right),\left( - \sqrt{2}, 2 \right)\]
\[\text{ Differentiating (1) w.r.t.x, }\]
\[\frac{dy}{dx} = - 2x . . . . . \left( 3 \right)\]
\[\text { Differentiating (2) w.r.t.x },\]
\[\frac{dy}{dx} = 2x . . . . . \left( 4 \right)\]
\[\text { Case } 1:\left( x, y \right)=\left( \sqrt{2}, 2 \right)\]
\[\text { From } \left( 3 \right), \text { we have,} m_1 = - 2\sqrt{2}\]
\[\text { From} \left( 4 \right) \text { we have }, m_2 = 2\sqrt{2}\]
\[\text { Now }, \]
\[\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{- 2\sqrt{2} - 2\sqrt{2}}{1 - 8} \right| = \frac{4\sqrt{2}}{7}\]
\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{4\sqrt{2}}{7} \right)\]
\[\text { Case } 1:\left( x, y \right)=\left( -\sqrt{2}, 2 \right)\]
\[\text { From } \left( 3 \right), \text { we have }, m_1 = 2\sqrt{2}\]
\[\text { From } \left( 4 \right) \text { we have }, m_2 = - 2\sqrt{2}\]
\[\text { Now,} \]
\[\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{2\sqrt{2} + 2\sqrt{2}}{1 - 8} \right| = \frac{4\sqrt{2}}{7}\]
\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{4\sqrt{2}}{7} \right)\]
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