If the normals at $A (t_{1})$ and $B (t_{2})$ meet again at $C (t_{3})$ on the parabola $y^{2} = 4 a x,$ then the locus of the mid point of $A B$ is

y^{2} = 2 a x - a^{2}

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y^{2} = 2 a x + 4 a^{2}

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y^{2} = 2 a x - 4 a^{2}

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y^{2} = 2 a x + a^{2}

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Solution

The correct option is B $y^{2} = 2 a x + 4 a^{2}$
If normal at $A (t_{1})$ meets again at $C (t_{3})$ on the parabola $y^{2} = 4 a x,$ then
$t_{3} = - t_{1} - \frac{2}{t_{1}} \dots (i)$

If normal at $B (t_{2})$ meets again at $C (t_{3})$ on the parabola, then
$t_{3} = - t_{2} - \frac{2}{t_{2}} \dots (i i)$

From equations $(i)$ and $(i i),$
$- t_{1} - \frac{2}{t_{1}} = - t_{2} - \frac{2}{t_{2}} \Rightarrow t_{1} t_{2} = 2 \dots (i i i)$

Let $(h, k)$ be the mid point of $A B,$ then
$h = \frac{a (t_{1}^{2} + t_{2}^{2})}{2}$
$\Rightarrow \frac{2 h}{a} = t_{1}^{2} + t_{2}^{2} \dots (i v)$
and $k = a (t_{1} + t_{2})$
$\Rightarrow \frac{k^{2}}{a^{2}} = t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2}$

From equations $(i i i)$ and $(i v),$ we get
$\frac{k^{2}}{a^{2}} = \frac{2 h}{a} + 4$
Hence, required locus is $y^{2} = 2 a x + 4 a^{2}$

Suggest Corrections

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