Find the condition that curves $2 x = y^{2}$ and 2xy = k inersect orthogonally.

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Solution

Given equation of curves are $2 x = y^{2} a n d 2 x y = k \Rightarrow y = \frac{k}{2 x} f r o m e q (i), 2 x = {(\frac{k}{2 x})}^{2} \Rightarrow 8 x^{3} = k^{2} \Rightarrow x^{3} = \frac{1}{8} k^{2} \Rightarrow x = \frac{1}{2} k^{\frac{2}{3}} ∴ y = \frac{k}{2 x} = \frac{k}{2 \frac{1}{2} k^{\frac{2}{3}}} = k^{\frac{1}{3}}$
From Eqs. (i) and (ii),
$2 = 2 y \frac{d y}{d x}$
and $2 [x . \frac{d y}{d x} + y . 1] = 0 \Rightarrow \frac{d y}{d x} = \frac{1}{y} a n d (\frac{d y}{d x}) = \frac{- 2 y}{2 x} = - \frac{y}{x} \Rightarrow {(\frac{d Y}{d x})}_{(\frac{1}{2} k^{\frac{2}{3}} k^{\frac{1}{3}})} = \frac{- k^{\frac{1}{3}}}{\frac{1}{2} K^{\frac{2}{3}}} = - 2 K^{- \frac{1}{3}}$
Since, the curves intersect orthogonally.
i.e., $m_{1} . m_{2} = - 1 \Rightarrow \frac{1}{k^{\frac{1}{3}}} . (- 2 k^{- \frac{1}{3}}) = - 1 \Rightarrow - 2 k^{- \frac{2}{3} = - 1} \Rightarrow \frac{2}{k^{\frac{2}{3}}} = 1 \Rightarrow k^{\frac{2}{3}} = 2 ∴ k^{2} = 8$

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