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Slope Point Form of a Line

If the normal...

Question

If the normal at point $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ on the parabola $y^{2} = 4 a x$ meet on the parabola, the $t_{1} t_{2} = 2$ .

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Solution

Equation of parabola $y^{2} = 4 a x$

And the normal point

$P (a {t_{1}}^{2}, 2 a t_{1})$

$Q (a {t_{2}}^{2}, 2 a t_{2})$

Prove that:-

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

Proof:-

We know that,

If normal at point $t_{1}$ to the parabola $y^{2} = 4 a x$ meets the parabola again at point $t_{3},$

Then,

$t_{3} = - t_{1} - \frac{2}{t_{1}} (Importent)$

According to given question,

Normals at point $t_{1}$ and $t_{2}$ meet at a point on parabola,

Hence, $t_{3}$ will be same for them,

$t_{3} = t_{3}$

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

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

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

$\Rightarrow 1 = \frac{2}{t_{1} t_{2}}$

$\Rightarrow t_{1} t_{2} = 2$

Hence proved.

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