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Question
Find the angle of intersection of the following curve x2 = 27y and y2 = 8x ?
Solution
\[\text { Given curves are },\]
\[ x^2 = 27y . . . \left( 1 \right)\]
\[ y^2 = 8x . . . \left( 2 \right)\]
\[\text { From } (2) \text { we get }\]
\[x = \frac{y^2}{8} \]
\[\text { Substituting this in }(1),\]
\[ \left( \frac{y^2}{8} \right)^2 = 27y\]
\[ \Rightarrow y^4 = 1728y\]
\[ \Rightarrow y \left( y^3 - {12}^3 \right) = 0\]
\[ \Rightarrow y = 0 ory = 12\]
\[\text { Substituting the values of y in (2), we get }, \]
\[ \Rightarrow x = 0 orx = 18\]
\[ \Rightarrow \left( x, y \right)=\left( 0, 0 \right),\left( 18, 12 \right)\]
\[\text { Differentiating (1) w.r.t.x },\]
\[2x = 27\frac{dy}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2x}{27} . . . \left( 3 \right)\]
\[\text { Differentiating (2) w.r.t.x },\]
\[2y \frac{dy}{dx} = 8\]
\[ \Rightarrow \frac{dy}{dx} = \frac{4}{y} . . . \left( 4 \right)\]
\[\text { Case } - 1:\left( x, y \right)=\left( 0, 0 \right)\]
\[\text { From }\left( 4 \right) \text { we have,} m_2 \text { is undefined }\]
\[ \therefore\text { We cannot find } \theta\]
\[\text { Case -} 2: \left( x, y \right)=\left( 18, 12 \right)\]
\[\text { From } \left( 3 \right) \text { we have }, m_1 = \frac{36}{27} = \frac{4}{3}\]
\[\text { From } \left( 4 \right) \text { we have }, m_2 = \frac{4}{12} = \frac{1}{3}\]
\[\text { Now }, \]
\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{9}} \right| = \frac{9}{13}\]
\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{9}{13} \right)\]
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