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Standard XII

Mathematics

Equation of Normal at Given Point

The chord AB ...

Question

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

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

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

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

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None of these

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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
$∴ \frac{4 a t_{1} + 2 a t_{2}}{3} = 0 \Rightarrow t_{2} + 2 t_{1} = 0$

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Similar questions

Q. The chord

A B

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

A C : A B = 1 : 3

then

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. 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

Q. If

P Q

is a normal chord of the parabola

y^{2} = 4 a x

P (a t^{2}, 2 a t) .

Then the axis of the parabola divides

¯ ¯¯¯¯¯¯ ¯ P Q

in the ratio

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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The chord AB of the parabola y2=4ax cuts the axis of the parabola at C. If A=(at21,2at1),B=(at22,2at2) and AC : AB = 1:3 then

The correct option is B t2+2t1=0 Here, C=(2at21+at223,4at1+2at23) It lies on y = 0 ∴4at1+2at23=0⇒t2+2t1=0

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

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
$∴ \frac{4 a t_{1} + 2 a t_{2}}{3} = 0 \Rightarrow t_{2} + 2 t_{1} = 0$