If the normal to $y^{2} = 4 ax$ at $t_{1}$ cuts the parabola again at $t_{2}$ then

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$- 8 \leq t_{2} \leq 8$

$t_{2}^{2} < 8$

$t_{2}^{2} \leq 8$

$t_{2}^{2} \geq 8$

answer is B.

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Detailed Solution

Given parabola $y^{2} = 4 ax$

Equation of normal at $t_{1}$ is $t_{1} x + y = 2 {at}_{1} + {at}_{1}^{3}$ it also meets parabola at $t_{2} ({at}^{2}_{2}, 2 {at}_{2})$ then $t_{1} ({at}_{2}^{2}) + 2 {at}_{2} = 2 {at}_{1} + {at}_{1}^{3} {at}_{1} {t_{2}}^{2} + 2 {at}_{2} = 2 {at}_{1} + {at}_{1}^{3} {at}_{1} {t_{2}}^{2} - {at}_{1}^{3} = 2 {at}_{1} - 2 {at}_{2} a t_{1} [{t_{2}}^{2} - {t_{1}}^{2}] = 2 a [t_{1} - t_{2}] - t_{1} [t_{1} + t_{2}] = 2 - {t_{1}}^{2} - t_{1} t_{2} = 2 {t_{1}}^{2} + t_{2} t_{1} + 2 = 0 I t i s a q u a d r a t i c e q u a t i o n F o r r e a l s o l u t i o n s, D \geq 0 \Rightarrow {t_{2}}^{2} - 4 (1) (2) \geq 0 \Rightarrow {t_{2}}^{2} \geq 8 O R n o r m a l a t t_{1} c u t s p a r a b o l a a g a i n a t t_{2} t h e n t_{2} = - t_{1} - \frac{2}{t_{1}} {t_{1}}^{2} + t_{2} t_{1} + 2 = 0 I t i s a q u a d r a t i c e q u a t i o n F o r r e a l s o l u t i o n s, D \geq 0 \Rightarrow {t_{2}}^{2} - 4 (1) (2) \geq 0 \Rightarrow {t_{2}}^{2} \geq 8$