Find the number of distinct normals that can be drawn from $(2, 1)$ to the parabola $y^{2} - 4 x - 2 y - 3 = 0$

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Solution

Given parabola is, $y^{2} - 4 x - 2 y - 3 = 0$
$\Rightarrow y^{2} - 2 y + 1 = 4 x + 4$
$\Rightarrow (y - 1)^{2} = 4 (x + 1) . . (1)$
Thus equation of normal to the parabola (1) is given by,
$(y - 1) + t (x + 1) = 2 t + t^{3}$
Also given that it passes through $(2, 1)$
$\Rightarrow 3 t = 2 t + t^{3} \Rightarrow t^{3} - t = 0 \Rightarrow t = 0, 1, - 1$
Hence, there can be three distinct normal drawn from $(2, 1)$ to the given parabola.

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