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, $A_1=\left(2t_1^2,4t_1\right)$ and $A_2=\left(2t_1^2,-4t_1\right)$
Then, the slope of $O{A}_1$ is $\frac{4t_1}{2t_1^2}=\frac{2}{t_1}$
$\Rightarrow tan30^{\circ }=\frac{2}{t_1}\Rightarrow t_1=2\sqrt{3}$
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,
$0=-2\sqrt{3}h+56\sqrt{3}$
$\Rightarrow h=28.$