Sub-normal to xy $=$ 9 at any point on it varies directly as

cube of ordinate

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square of ordinate

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ordinate

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cube of abscissa

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Solution

The correct option is A cube of ordinate
$S N = (y m)$
$y^{'} = \frac{- 9}{x^{2}}$
$S N = (\frac{y - 9}{x^{2}})$
$S N = (\frac{- y^{3}}{9})$
$S N α y^{3}$

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