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Conic Section

If xyy-x=16, ...

Question

If $x y (y - x) = 16,$ then $y$ has a stationary point when $x$ is equal to

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Solution

$x y (y - x) = 16 \dots (1)$
$\Rightarrow x y^{2} - x^{2} y = 16$
Differentiate w.r.t. $x$
$\Rightarrow 2 x y y^{'} + y^{2} - x^{2} y^{'} - 2 x y = 0$
$\Rightarrow y^{'} = \frac{2 x y - y^{2}}{2 x y - x^{2}} = \frac{y (2 x - y)}{x (2 y - x)}$
For stationary point $y^{'} = 0$
Only,
$\Rightarrow 2 x = y \dots (2)$
as $y \neq 0$
By solving $(1)$ and $(2)$
$2 x^{2} (2 x - x) = 16$ , we get
$\Rightarrow 2 x^{3} = 16 \Rightarrow x = 2$

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