Question

Find the angle of intersection of the following curve  x² = 27y and y² = 8x ?

Answer 1

\[\text {  Given curves are },\]

\[ x^2 = 27y . . . \left( 1 \right)\]

\[ y^2 = 8x . . . \left( 2 \right)\]

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

\[x = \frac{y^2}{8} \]

\[\text { Substituting this in  }(1),\]

\[ \left( \frac{y^2}{8} \right)^2 = 27y\]

\[ \Rightarrow y^4 = 1728y\]

\[ \Rightarrow y \left( y^3 - {12}^3 \right) = 0\]

\[ \Rightarrow y = 0 ory = 12\]

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

\[ \Rightarrow x = 0 orx = 18\]

\[ \Rightarrow \left( x, y \right)=\left( 0, 0 \right),\left( 18, 12 \right)\]

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

\[2x = 27\frac{dy}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{2x}{27} . . . \left( 3 \right)\]

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

\[2y \frac{dy}{dx} = 8\]

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

\[\text { Case } - 1:\left( x, y \right)=\left( 0, 0 \right)\]

\[\text { From  }\left( 4 \right) \text { we have,} m_2 \text { is undefined }\]

\[ \therefore\text { We cannot find } \theta\]

\[\text { Case -} 2: \left( x, y \right)=\left( 18, 12 \right)\]

\[\text { From } \left( 3 \right) \text { we have }, m_1 = \frac{36}{27} = \frac{4}{3}\]

\[\text { From } \left( 4 \right) \text { we have }, m_2 = \frac{4}{12} = \frac{1}{3}\]

\[\text { Now }, \]

\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{9}} \right| = \frac{9}{13}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{9}{13} \right)\]