Question

A normal with slope \(\frac{1}{\sqrt6}\) is drawn from the point (0, -α) to the parabola x² = -4αy, where a > 0. Let L be the line passing through (0, -α) and parallel to the directrix of the parabola. Suppose that L intersects the parabola at two points A and B. Let r denote the length of the latus rectum and s denote the square of the length of the line segment AB. If r : s = 1 : 16 then the value of 24a is __________.

Answer 1

Correct Answer: 12

Equation of the Normal and Intersection Points 

The equation of the normal to \( x^2 = -4ay \) is:

\[ -ty + x = 2at + at^3 \]

With slope \( 1 / \sqrt{6} \), we have \( t = \sqrt{6} \).

Substituting \( t = \sqrt{6} \):

\[ y = -2a - at^2 - 8a = -8a \quad \Rightarrow \quad \alpha = 8a \]

Points of Intersection A and B:

\[ A = (\alpha \sqrt{2}, -\alpha), \quad B = (-\alpha \sqrt{2}, -\alpha) \]

Length of AB:

\[ AB = \sqrt{(\alpha \sqrt{2} - (-\alpha \sqrt{2}))^2 + (-\alpha - (-\alpha))^2} \] Simplifying: \[ AB = 2\alpha \sqrt{2} \]

Given:

\[ r = 4a, \quad s = (AB)^2 = 2\alpha = 128a^2 \]

Solving:

\[ \frac{r}{s} = \frac{1}{32a} = \frac{1}{16} \quad \Rightarrow \quad a = \frac{1}{2} \]

Thus, \( 24a = 12 \).