Question

Find the angle of intersection of the following curve \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] and x² + y² = ab ?

Answer 1

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

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 . . . \left( 1 \right)\]

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

\[\text { Multiplying } (2) by\frac{1}{a^2},\]

\[\frac{x^2}{a^2} + \frac{y^2}{a^2} = \frac{b}{a} . . . \left( 3 \right)\]

\[\text { Subtracting (1) from (3), we get }\]

\[\frac{y^2}{a^2} - \frac{y^2}{b^2} = \frac{b}{a} - 1\]

\[ \Rightarrow y^2 \left( \frac{b^2 - a^2}{a^2 b^2} \right) = \frac{b - a}{a}\]

\[ \Rightarrow y^2 = \frac{b - a}{a} \times \frac{a^2 b^2}{\left( b + a \right)\left( b - a \right)} = \frac{a b^2}{b + a}\]

\[ \Rightarrow y = \pm b\sqrt{\frac{a}{b + a}}\]

\[\text { Substituting this in } (3),\]

\[\frac{x^2}{a^2} + \frac{a b^2}{\left( b + a \right)\left( a^2 \right)} = \frac{b}{a}\]

\[ \Rightarrow \left( a + b \right) x^2 + a b^2 = a b^2 + a^2 b\]

\[ \Rightarrow x^2 = \frac{a^2 b}{a + b}\]

\[ \Rightarrow x = \pm a\sqrt{\frac{b}{a + b}}\]

\[ \therefore \left( x, y \right)=\left( \pm a\sqrt{\frac{b}{a + b}}, \pm b\sqrt{\frac{a}{b + a}} \right)\]

\[\text { Now },\left( x, y \right)=\left( a\sqrt{\frac{b}{a + b}}, b\sqrt{\frac{a}{b + a}} \right)\]

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

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

\[ \Rightarrow \frac{dy}{dx} = \frac{- x b^2}{a^2 y}\]

\[ \Rightarrow m_1 = \frac{- a b^2 \sqrt{\frac{b}{a + b}}}{a^2 b\sqrt{\frac{a}{b + a}}} = \frac{- b\sqrt{b}}{a\sqrt{a}}\]

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

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

\[ \Rightarrow \frac{dy}{dx} = \frac{- x}{y}\]

\[ \Rightarrow m_2 = \frac{- a\sqrt{\frac{b}{a + b}}}{b\sqrt{\frac{a}{b + a}}} = \frac{- a\sqrt{b}}{b\sqrt{a}}\]

\[\text { We have,} \]

\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{- b\sqrt{b}}{a\sqrt{a}} + \frac{a\sqrt{b}}{b\sqrt{a}}}{1 + \left( \frac{b\sqrt{b}}{a\sqrt{a}} \right)\left( \frac{a\sqrt{b}}{b\sqrt{a}} \right)} \right| = \frac{\frac{- b^2 \sqrt{ab} + a^2 \sqrt{ab}}{a^2 b}}{\frac{a^2 b + a b^2}{a^2 b}} = \frac{\sqrt{ab}\left( a + b \right)\left( a - b \right)}{a^2 b} \times \frac{a^2 b}{ab\left( a + b \right)} = \frac{a - b}{\sqrt{ab}}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{a - b}{\sqrt{ab}} \right)\]

\[ {\text { Similarly, we can prove that }\theta=tan}^{- 1} \left( \frac{a - b}{\sqrt{ab}} \right) \text { for all possibilities of } \left( x, y \right)\]