The straight line which is both tangent and normal to the curve $x = 3 t^{2}, y = 2 t^{3}$ is:

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$y + \sqrt{2} (x - 2) = 0$

$y - \sqrt{3} (x - 1) = 0$

$y + \sqrt{3} (x - 1) = 0$

$y - \sqrt{2} (x - 2) = 0$

answer is C, D.

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Detailed Solution

Let line is tangent at $(3 t_{1}^{2}, 2 t_{1}^{3})$ and normal at $(3 t_{2}^{2}, 2 t_{2}^{3})$
$\Rightarrow {\frac{d y}{d x} |}_{3 t_{1}^{2}, 2 t_{1}^{3}} = t_{1}$
So, slope of normal at $(3 t_{2}^{2}, 2 t_{2}^{3}) = - \frac{1}{t_{2}}$
$\Rightarrow t_{1} = - \frac{1}{t_{2}} t_{2} = - \frac{1}{t_{1}} \Rightarrow t_{1} = \frac{2 t_{1}^{3} + \frac{2}{t_{1}^{3}}}{3 (t_{1}^{2} - \frac{1}{t_{1}^{2}})} \Rightarrow t_{1}^{2} (t_{1}^{4} - 3) = 2 o r t_{1} \pm \sqrt{2}$