The equations of the tangents at the origin to the curve $y^{2} = x^{2} (1 + x)$ are

y = \pm x

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y = \pm 2 x

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y = \pm 3 x

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x = \pm 2 y

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Solution

The correct option is C $y = \pm x$
$y = +_{-} \sqrt{x^{2} + x^{3}}$
$y^{'} = +_{-} \frac{2 x + 3 x^{2}}{2 \sqrt{x^{2} + x^{3}}}$
$y^{'} = +_{-} \frac{2 + 3 x}{2 \sqrt{1 + x}}$
$y^{'} |_{x = 0} = +_{-} 1$
$y = +_{-} x$

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