Normals $A O, A A_{1}, A A_{2}$ are drawn to parabola $y^{2} = 8 x$ from the point $A (h, 0)$ . If triangle $O A_{1} A_{2}$ is equilateral, then possible value of $h$ is

26

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24

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28

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none of these

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Solution

The correct option is C $28$
Given equation of parabola is
$y^{2} = 8 x$
Here, $a = 2$
Let, $A_{1} \equiv (2 t_{1}^{2}, 4 t_{1}), A_{2} \equiv (2 t_{1}^{2}, - 4 t_{1})$
Slope of $O A_{1} = \frac{4 t_{1}}{2 t_{1}^{2}} = \frac{2}{t_{1}}$
Since, $O A_{1} A_{2}$ is equilateral triangle , having all angles $\frac{π}{3}$
Clearly, $∠ A_{1} O A = \frac{π}{6}$
$\Rightarrow \frac{2}{t_{1}} = \frac{1}{\sqrt{3}}$
$\Rightarrow t_{1} = 2 \sqrt{3}$
Equation of normal at $A_{1}$ is $y = - t_{1} x + 4 t_{1} + 2 t_{1}^{3}$
Since, it passes through $(h, 0)$
$\Rightarrow h = 4 + 2 t_{1}^{2} = 4 + 2 (12) = 28$

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