Question

Find the angle of intersection of the following curve y = 4 − x² and y = x² ?

Answer 1

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

\[y = 4 - x^2 . . . . . \left( 1 \right)\]

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

\[\text { From ( 1)and (2), we get }\]

\[4 - x^2 = x^2 \]

\[ \Rightarrow 2 x^2 = 4\]

\[ \Rightarrow x^2 = 2\]

\[ \Rightarrow x = \pm \sqrt{2}\]

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

\[ \Rightarrow y = 2\]

\[ \Rightarrow \left( x, y \right)=\left( \sqrt{2},2 \right),\left( - \sqrt{2}, 2 \right)\]

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

\[\frac{dy}{dx} = - 2x . . . . . \left( 3 \right)\]

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

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

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

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

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

\[\text { Now }, \]

\[\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{- 2\sqrt{2} - 2\sqrt{2}}{1 - 8} \right| = \frac{4\sqrt{2}}{7}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{4\sqrt{2}}{7} \right)\]

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

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

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

\[\text { Now,} \]

\[\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{2\sqrt{2} + 2\sqrt{2}}{1 - 8} \right| = \frac{4\sqrt{2}}{7}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{4\sqrt{2}}{7} \right)\]