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

Q. The normal a point

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

on a parabola meets the parabola again in the point

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

then

Q. Match the column
The parabola

y^{2} = 4 a x

has a chord

A B

joining points

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

and

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

Column - I	Column - II
(A) $A B$ is a normal chord if	(P) $t_{2} = - t_{1} - \frac{1}{2}$
(B) $A B$ is a focal chord	(Q) $t_{2} = \frac{- 4}{t_{2}}$
(C) $A B$ subtends $90^{\circ}$ at point $(0, 0)$ if	(R) $t_{2} = \frac{- 1}{t_{1}}$ (S) $t_{2} = - t_{1} - \frac{2}{t_{1}}$

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t1 and t2 are two points on the parabola y2=4x. If the chord joining them is a normal to the parabola at t1 then

The correct option is C t1(t1+t2)+2=0Normal at x=at21,y=2at1 is given by,y+t1x=2at1+at31Given it also passes through ′t′2) which is x=at22,y=2at2⇒2at2+at22t1=2at1+at31⇒2(t2−t1)=t1(t21−t22)=t1(t1+t2)(t1−t2)⇒2=−t1(t1+t2) (since t1≠t2)⇒t1(t1+t2)+2=0

$t_{1}$ and $t_{2}$ are two points on the parabola $y^{2} = 4 x$ . If the chord joining them is a normal to the parabola at $t_{1}$ then