$y^{2} = 16 x$ and $2 x^{2} + y^{2} = 4$ ,then show that the curves cut each other orthogonally.

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Solution

$y^{2} = 16 x$
$2 y \frac{d y}{d x} = 16$
$∴ {(\frac{d y}{d x})}_{I} = \frac{8}{y} = m_{1}$

$2 x^{2} + y^{2} = 4.$
$4 x + 2 y \frac{d y}{d x} = 0$
$∴ {(\frac{d y}{d x})}_{I I} = \frac{- 2 x}{y} = - \frac{y}{8} = m_{2}$
Clearly $m_{1} m_{2} = - 1$
Hence, the curves cut orthogonally.

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