Question

Find the angle of intersection of the following curve  y = x² and x² + y² = 20  ?

Answer 1

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

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

\[ x^2 + y^2 = 20 . . . \left( 2 \right)\]

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

\[y + y^2 = 20\]

\[ \Rightarrow y^2 + y - 20 = 0\]

\[ \Rightarrow \left( y + 5 \right)\left( y - 4 \right) = 0\]

\[ \Rightarrow y = - 5 ory = 4\]

\[\text { Substituting the values of y in } \left( 1 \right) \text { we get }, \]

\[ x^2 = - 5 \text { or} x^2 = 4 \]

\[ \Rightarrow x = \pm 2 \text { and } x^2 = -5 \text { has no real solution }\]

\[\text { So },\left( x, y \right)=\left( 2, 4 \right)or \left( - 2, 4 \right)\]

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

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

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

\[2x + 2y \frac{dy}{dx} = 0\]

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

\[\text { Case  }-1:\left( x, y \right)=\left( 2, 4 \right)\]

\[\text { From } \left( 3 \right) \text { we have }, m_1 = 2\left( 2 \right) = 4\]

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

\[\text { Now }, \]

\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{4 + \frac{1}{2}}{1 + 4 \left( \frac{- 1}{2} \right)} \right| = \frac{9}{2}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{9}{2} \right)\]

\[\text { Case} -2:\left( x, y \right)=\left( - 2, 4 \right)\]

\[\text { From } \left( 3 \right) \text {we have,} m_1 = 2\left( - 2 \right) = - 4\]

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

\[\text { Now }, \]

\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{- 4 - \frac{1}{2}}{1 - 4 \left( \frac{1}{2} \right)} \right| = \frac{9}{2}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{9}{2} \right)\]