Question

Show that the curves 4x = y² and 4xy = k cut at right angles, if k² = 512 ?

Answer 1

\[\text { Given }: \]

\[4x = y^2 . . . \left( 1 \right)\]

\[4xy = k . . . \left( 2 \right)\]

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

\[ y^3 = k\]

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

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

\[4x = k^\frac{2}{3} \]

\[ \Rightarrow x = \frac{k^\frac{2}{3}}{4}\]

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

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

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

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( \frac{k^\frac{2}{3}}{4}, k^\frac{1}{3} \right) = \frac{2}{k^\frac{1}{3}} = 2 k^\frac{- 1}{3} \]

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

\[4x\frac{dy}{dx} + 4y = 0\]

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

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( \frac{k^\frac{2}{3}}{4}, k^\frac{1}{3} \right) = \frac{- k^\frac{1}{3}}{\left( \frac{k^\frac{2}{3}}{4} \right)} = - 4 k^\frac{- 1}{3} \]

\[\text { It is given that the curves intersect at right angles }.\]

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

\[ \Rightarrow 2 k^\frac{- 1}{3} \times - 4 k^\frac{- 1}{3} = - 1\]

\[ \Rightarrow 8 k^\frac{- 2}{3} = 1\]

\[ \Rightarrow k^\frac{- 2}{3} = \frac{1}{8}\]

\[ \Rightarrow k^\frac{2}{3} = 8\]

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

\[ k^2 = 512\]