Parabola
Parabola
nta_abhyas_2025
Grade None
Question:
Let points $A_1$, $A_2$ and $A_3$ lie on the parabola $y^2 = 8x$. If $A_1 A_2 A_3$ is an equilateral triangle and normals at points $A_1$, $A_2$ and $A_3$ on this parabola meet at the point $(h, 0)$, then the value of $h$ is
Step-by-Step Solution
Key Concept: Use the normal form for parabola and apply symmetry; the normals at parameters $t$ and $-t$ are symmetric about the axis.
One normal from $(h, 0)$ on parabola $y^2 = 8x$ is the $x$-axis itself, giving one vertex at origin. For the other two normals with parameters $t_1$ and $-t_1$ (by symmetry), the slope of $OA_1$ where $A_1 = (2t_1^2, 4t_1)$ equals $\frac{4t_1}{2t_1^2} = \frac{2}{t_1}$. Given $\tan 30° = \frac{1}{\sqrt{3}} = \frac{2}{t_1}$, we get $t_1 = 2\sqrt{3}$. The normal equation at parameter $t$ is $y = -tx + 4t + 2t^2$. Substituting $(h, 0)$ and $t_1 = 2\sqrt{3}$ yields $0 = -2\sqrt{3}h + 8\sqrt{3} + 24$, solving to $h = 28$.
Correct Answer: 28