Question

Show that the curves 2x = y² and 2xy = k cut at right angles, if k² = 8 ?

Answer 1

\[\text { Given }: \]

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

\[2xy = k . . . \left( 2 \right)\]

\[\text { From (1) and (2), we get }\]

\[ y^3 = k\]

\[ \Rightarrow y = k^\frac{1}{3} \]

\[\text { From (1), we get }\]

\[2x = k^\frac{2}{3} \]

\[ \Rightarrow x = \frac{k^\frac{2}{3}}{2}\]

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

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

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{y}\]

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( \frac{k^\frac{2}{3}}{2}, k^\frac{1}{3} \right) = \frac{1}{k^\frac{1}{3}} = k^\frac{- 1}{3} \]

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

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

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

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( \frac{k^\frac{2}{3}}{2}, k^\frac{1}{3} \right) = \frac{- k^\frac{1}{3}}{\left( \frac{k^\frac{2}{3}}{2} \right)} = - 2 k^\frac{- 1}{3} \]

\[\text { It is given that the curves intersect orthogonally }.\]

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

\[ \Rightarrow k^\frac{- 1}{3} \times - 2 k^\frac{- 1}{3} = - 1\]

\[ \Rightarrow 2 k^\frac{- 2}{3} = 1\]

\[ \Rightarrow k^\frac{- 2}{3} = \frac{1}{2}\]

\[ \Rightarrow k^\frac{2}{3} = 2\]

\[\text { Cubing on both sides, we get }\]

\[ k^2 = 8\]