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Perpendicular from the Center to a Chord Bisects the Chord

If t1 and ...

Question

If $t_{1}$ and $t_{2}$ are two points on the parabola $y^{2} = 4 a x$ . If the focal chord joining them coincides with the normal chord, then

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

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t_{1} + t_{2} = 0

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t_{1} t_{2} = - 1

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None of these

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Solution

The correct option is D None of these
Since, $t_{1}$ and $t_{2}$ are end-points of focal chord
$\Rightarrow t_{1} t_{2} = - 1$
or $t_{2} = - \frac{1}{t_{1}}$ .....(i)
Since, $t_{1}$ and $t_{2}$ are end-points of normal chord.
$\Rightarrow t_{2} = - t_{1} - \frac{2}{t_{1}}$ ......(ii)
From (i) and (ii)
$- \frac{1}{t_{1}} = - t_{1} - \frac{2}{t_{1}}$
$\Rightarrow {t_{1}}^{2} = - 1$
or $t_{1} = - i$
So no such chord is possible.

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Similar questions

Q. If the chord joining the points

t_{1}

and

t_{2}

to the parabola

y^{2} = 4 a x

is normal to the parabola at

t_{1}

then prove that

t_{1} (t_{1} + t_{2}) = - 2

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

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

Q. If

t_{1}

and

t_{2}

are the extremities of any focal chord of the parabola

y^{2} = 4 a x

then

t_{1} t_{2} =

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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If t1 and t2 are two points on the parabola y2=4ax. If the focal chord joining them coincides with the normal chord, then

The correct option is D None of theseSince, t1 and t2 are end-points of focal chord⇒t1t2=−1or t2=−1t1 .....(i)Since, t1 and t2 are end-points of normal chord.⇒t2=−t1−2t1 ......(ii)From (i) and (ii)−1t1=−t1−2t1 ⇒t12=−1or t1=−iSo no such chord is possible.

If $t_{1}$ and $t_{2}$ are two points on the parabola $y^{2} = 4 a x$ . If the focal chord joining them coincides with the normal chord, then