Question

Find the condition for the following set of curve to intersect orthogonally \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text { and } xy = c^2\] ?

Answer 1

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

\[xy = c^2 . . . \left( 2 \right)\]

\[\text { Let the curves intersect orthogonally at }\left( x_1 , y_1 \right).\]

\[\text { On differentiating (1) on both sides w.r.t.x, we get }\]

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

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

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = \frac{x_1 b^2}{a^2 y_1}\]

\[\text { On differentiating (2) on both sides w.r.t.x, we get }\]

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

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

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = \frac{- y_1}{x_1}\]

\[\text { It is given that the curves intersect orhtogonally at }\left( x_1 , y_1 \right).\]

\[ \therefore m_1 \times m_2 = - 1\]

\[ \Rightarrow \frac{x_1 b^2}{a^2 y_1} \times \frac{- y_1}{x_1} = - 1\]

\[ \Rightarrow a^2 = b^2 \]