Question

Let points $A_{1},A_{2}$ and $A_{3}$ lie on the parabola $y^{2}=8x.$ If $\Delta A_{1}A_{2}A_{3}$ is an equilateral triangle and normals at points $A_{1},A_{2}$ and $A_{3}$ on this parabola meet at the point $\left(h , 0\right)$ , then the value of $h$ is

• A. $24$
• B. $26$
• C. $38$
• D. $28$

Answer 1

Answer: (D)

One of the normal from $\left(h , 0\right)$ on the parabola $y^{2}=8x$ is $y=0$
Hence, assume $A_{3}=\left(0 , 0\right)$
Let, $A_{1}=\left(2 t_{1}^{2} , 4 t_{1}\right)$ and $A_{2}=\left(2 t_{1}^{2} , - 4 t_{1}\right)$
Then, the slope of $O A_{1}$ is $\frac{4 t_{1}}{2 t_{1}^{2}}=\frac{2}{t_{1}}$
$\Rightarrow tan30^\circ =\frac{2}{t_{1}}\Rightarrow t_{1}=2\sqrt{3}$
The equation of the normal at $A_{1}$ is
$\Rightarrow y=-2\sqrt{3}x+8\sqrt{3}+48\sqrt{3}$
Putting $\left(h , 0\right)$ in the above equation, we get,
$0=-2\sqrt{3}h+56\sqrt{3}$
$\Rightarrow h=28.$