Question

Find the angle of intersection of the following curve  2y² = x³ and y² = 32x ?

Answer 1

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

\[2 y^2 = x^3 . . . \left( 1 \right)\]

\[ y^2 = 32x . . . \left( 2 \right) \]

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

\[2\left( 32x \right) = x^3 \]

\[ \Rightarrow 64x = x^3 \]

\[ \Rightarrow x\left( x^2 - 64 \right) = 0\]

\[ \Rightarrow x = 0, 8 , - 8\]

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

\[ y_1 = 0, 16, - 16\]

\[ \therefore \left( x_1 , y_1 \right)=\left( 0, 0 \right),\left( 8, 16 \right) or \left( 8, - 16 \right) \]

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

\[4y \frac{dy}{dx} = 3 x^2 \]

\[ \Rightarrow \frac{dy}{dx} = \frac{3 x^2}{4y} . . . \left( 3 \right)\]

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

\[2y\frac{dy}{dx} = 32\]

\[ \Rightarrow \frac{dy}{dx} = \frac{16}{y} . . . \left( 4 \right)\]

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

\[\text { From }\left( 3 \right) \text { we have, } m_1 = \frac{0}{0} \]

\[ \therefore\text {  We cannot determine theta in this case }.\]

\[\text { Case}-2:\left( x, y \right)=\left( 8, 16 \right)\]

\[\text { From }\left( 3 \right) \text { we have,} m_1 = \frac{192}{64} = 3\]

\[\text { From} \left( 4 \right) \text { we have,} m_2 = \frac{16}{16} = 1\]

\[\text { Now,} \]

\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{3 - 1}{1 + 3} \right| = \frac{2}{4} = \frac{1}{2}\]

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

\[\text { Case}- 3:\left( x_1 , y_1 \right)=\left( 8, - 16 \right)\]

\[\text { From } \left( 3 \right) \text { we have }, m_1 = \frac{192}{- 64} = - 3\]

\[\text { From } \left( 4 \right) \text { we have }, m_2 = \frac{16}{- 16} = - 1\]

\[\text { Now, } \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{- 3 + 1}{1 + 3} \right| = \frac{2}{4} = \frac{1}{2}\]

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