Any chord $A B o f y 62 = 4 a x$ cuts the axis of the parabola at $P$ , so that $A P : A B = 1 : 3 w h e r e A (a t_{1}^{2}, 2 a t_{1}), B (a t_{2}^{2}, 2 a t_{2})$ , then

2 t_{1} + t_{2} = 0

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t_{1} + 2 t_{2} = 0

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t_{1} t_{2} = - 4

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t_{1} + t_{2} = 0

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Solution

The correct option is A $2 t_{1} + t_{2} = 0$
$A P : A B = 1 : 3$

$& t h e r e 4; A P : P B = 1 : 2$

$& t h e r e 4; P has coordinates$

$(\frac{a t_{2}^{2} + 2. a t_{1}^{2}}{3}, \frac{2 a t_{2} + 2.2 a t_{1}}{3})$

$P$ also lies on axis of parabola i.e., $y = 0$

$\Rightarrow 2 t_{1} + t_{2} = 0$

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Any chord AB of y62=4ax cuts the axis of the parabola at P, so that AP:AB=1:3 where A(at21, 2at1), B(at22, 2at2), then

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