If three distinct and real normals can be drawn to $y^{2} = 8 x$ from the point $(a, 0)$ , then-

a > 2

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a \in (2, 4)

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a > 4

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none of these

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Solution

The correct option is D $a > 4$
Equation of normal to the parabola is given by,
$y + t x = 2 a t + a t^{3}$
Here $a = 2$ so,
$y + t x = 4 t + 2 t^{3}$
Given it passes through $(a, 0)$
$\Rightarrow t a = 4 t + 2 t^{3} \Rightarrow t (2 t^{2} - a + 4) . . (1)$
Therefore for three distinct normal through $(a, 0)$
equation (1) must have three distinct roots
$\Rightarrow - a + 4 < 0 \Rightarrow a > 4$

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