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Definition of a Parabola

The normal at...

Question

The normal at the 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

t_{2} = - t_{1} - \frac{2}{t_{1}}

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t_{2} = - t_{1} + \frac{2}{t_{1}}

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t_{2} = t_{1} - \frac{2}{t_{1}}

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

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Solution

The correct option is A $t_{2} = - t_{1} - \frac{2}{t_{1}}$
$y^{2} = 4 b x$
$\Rightarrow \frac{d y}{d x} = \frac{2 b}{y}$
Therefore, at the point $(b t_{1}^{2}, 2 b t_{1}), \frac{d y}{d x} = \frac{1}{t_{1}}$

So, the slope of normal at $(b t_{1}^{2}, 2 b t_{1}) = \frac{- 1}{(\frac{d y}{d x})} = - t_{1}$

Therefore, the equation of normal at $(b t_{1}^{2}, 2 b t_{1})$ is $(y - 2 b t_{1}) = - t_{1} (x - b t_{1}^{2})$

The point $(b t_{2}^{2}, 2 b t_{2})$ also lies on the normal
$⟹ (2 b t_{2} - 2 b t_{1}) = - t_{1} (b t_{2}^{2} - b t_{1}^{2})$
$⟹ 2 b (t_{2} - t_{1}) = - t_{1} b (t_{2} - t_{1}) (t_{2} + t_{1})$
$⟹ 2 b (t_{2} - t_{1}) + t_{1} b (t_{2} - t_{1}) (t_{2} + t_{1}) = 0$
$⟹ b (t_{2} - t_{1}) (2 + t_{1} (t_{2} + t_{1})) = 0$
$⟹ 2 + t_{1} (t_{2} + t_{1}) = 0$ .... $[t_{2} \neq t_{1}]$ and [ $b \neq 0$ ]
$⟹ t_{2} = - t_{1} - \frac{2}{t_{1}}$
The answer is option (A).

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