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The normal at...

Question

The normal at $t_{1}$ on the parabola $y^{2} = 4 a x$ meets the parabola at $t_{2}$ then $(t_{1} + \frac{2}{t_{1}})$ is:

A

- t_{2}

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B

t_{2}

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C

t_{1} + t_{2}

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D

\frac{1}{t_{2}}

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Solution

The correct option is A $- t_{2}$
Equation of the parabola is $y^{2} = 4 a x$

Equation of normal at $t_{1} = (a t_{1}^{2}, 2 a t_{1})$ is $y + x t_{1} = 2 a t_{1} + a t_{1}^{3}$ .

This normal meets the parabola again at $(a t_{2}^{2}, 2 a t_{2})$ .

Therefore, $2 a t_{2} + a t_{2}^{2} t_{1} = 2 a t_{1} + a t_{1}^{3}$

$\Rightarrow 2 (t_{2} - t_{1}) = t_{1} (t_{1}^{2} - t_{2}^{2})$

$\Rightarrow 2 = - t_{1} (t_{1} + t_{2})$

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

Divide by $t_{1}$ we get

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

$∴ t_{1} + \frac{2}{t_{1}} = - t_{2}$

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Similar questions

Q. If the normal at

t_{1}

on the parabola

y^{2} = 4 a x

meets it again at

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t_{2} =

Q. The normal a point

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Q. If the chord joining the points

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t_{2}

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Q. Match the column
The parabola

y^{2} = 4 a x

A B

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

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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