Question

A normal with slope \( \frac{1}{\sqrt{6}} \) is drawn from the point \( (0, -\alpha) \) to the parabola \( x^2 = -4ay \), where \( a>0 \). Let \( L \) be the line passing through \( (0, -\alpha) \) 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 \_\_\_\_.

Hint:

Use parametric equations of conics to derive intersection points for normals and tangents.

Answer 1

The equation of the normal to \( x^2 = -4ay \) is: \[ -ty + x = 2at + at^3, \quad \text{with slope } \frac{1}{\sqrt{6}} \Rightarrow t = \sqrt{6}. \] Substituting \( t = \sqrt{6} \): \[ y = -2a - at^2 - 8a = -8a \quad \Rightarrow \quad \alpha = 8a. \] Points of intersection \( A \) and \( B \): \[ A = \left(\frac{\alpha}{\sqrt{2}}, -\alpha\right), \quad B = \left(-\frac{\alpha}{\sqrt{2}}, -\alpha\right). \] Length of \( AB \): \[ AB = \sqrt{\frac{2\alpha^2}{2}} = \sqrt{2\alpha}. \] Given \( r = 4a \), \( s = (AB)^2 = 2\alpha = 128a^2 \). Solving: \[ \frac{r}{s} = \frac{1}{32a} = \frac{1}{16} \quad \Rightarrow \quad a = \frac{1}{2}, \quad 24a = 12. \]