The chord AB of the parabola $y^{2} = 4 a x$ cuts the axis of the parabola at C. If $A = (a t_{1}^{2}, 2 a t_{1}), B = (a t_{2}^{2}, 2 a t_{2})$ and AC : AB = 1:3 then

$t_{2} = 2 t_{1}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$t_{2} + 2 t_{1} = 0$

Here, $C = (\frac{2 a t_{1}^{2} + a t_{2}^{2}}{3}, \frac{4 a t_{1} + 2 a t_{2}}{3})$
It lies on y = 0
$∴ \frac{4 a t_{1} + 2 a t_{2}}{3} = 0 \Rightarrow t_{2} + 2 t_{1} = 0$

Suggest Corrections

Similar questions

The chord AB of the parabola $y^{2} = 4 a x$ cuts the axis of the parabola at C. If $A = (a t_{1}^{2}, 2 a t_{1}), B = (a t_{2}^{2}, 2 a t_{2})$ and AC : AB = 1:3 then

Q. If the coordinates of the Parabola of the ends of a focal chord of Parabola

y^{2} = 4 a x

are

(a t_{1}^{2}, 2 a t_{1})

and

(a t_{2}^{2}, 2 a t_{2})

, then prove that :

t_{1} . t_{2} = - 1

If the chord joining the points $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ of the parabola $y^{2} = 4 a x$ passes through the focus of the parabola, then