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, \( 24a = 12 \).

Answer 2

To solve the problem, let's analyze the parabola and the given lines, and use the given ratios to find the value of \( 24a \).

Given:
- Parabola: \( x^2 = -4 a y \), where \( a > 0 \).
- Point: \( (0, -\alpha) \).
- A normal with slope \( \frac{1}{\sqrt{6}} \) is drawn from \( (0, -\alpha) \) to the parabola.
- Line \( L \) passes through \( (0, -\alpha) \) and is parallel to the directrix of the parabola.
- \( L \) intersects the parabola at points \( A \) and \( B \).
- \( r \) = length of the latus rectum.
- \( s \) = square of length of segment \( AB \).
- Ratio \( r : s = 1 : 16 \).
- Find \( 24a \).

Step 1: Understand the parabola
The parabola \( x^2 = -4 a y \) opens downward.
- Focus: \( (0, -a) \)
- Directrix: \( y = a \)
- Length of latus rectum: \( r = 4a \).

Step 2: Equation of line \( L \)
- Since \( L \) is parallel to directrix, \( L \) is a horizontal line.
- Passing through \( (0, -\alpha) \) means:
\[ L: y = -\alpha \]

Step 3: Find points \( A \) and \( B \) where \( L \) intersects parabola
- Substitute \( y = -\alpha \) in parabola:
\[ x^2 = -4 a (-\alpha) = 4 a \alpha \] - So the points are:
\[ A = (-2 \sqrt{a \alpha}, -\alpha), \quad B = (2 \sqrt{a \alpha}, -\alpha) \]

Step 4: Calculate length \( AB \)
\[ AB = 2 \times 2 \sqrt{a \alpha} = 4 \sqrt{a \alpha} \] \[ s = (AB)^2 = (4 \sqrt{a \alpha})^2 = 16 a \alpha \]

Step 5: Use the slope of the normal
- Equation of parabola in parametric form:
\[ x = 2 a t, \quad y = -a t^2 \] - Slope of tangent at point \( t \):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2 a t}{2 a} = -t \] - Slope of normal:
\[ m = \frac{1}{t} \] - Given slope of normal is \( \frac{1}{\sqrt{6}} \), so:
\[ t = \sqrt{6} \] - Coordinates of point on parabola where normal is drawn:
\[ x_1 = 2 a t = 2 a \sqrt{6}, \quad y_1 = -a t^2 = -6 a \]

Step 6: Equation of normal passing through \( (0, -\alpha) \) and \( (x_1, y_1) \)
- Slope is \( \frac{1}{\sqrt{6}} \). Equation:
\[ y + \alpha = \frac{1}{\sqrt{6}} (x - 0) \] - Substituting \( (x_1, y_1) \):
\[ -6 a + \alpha = \frac{1}{\sqrt{6}} \times 2 a \sqrt{6} = 2 a \] - Rearranged:
\[ \alpha = 6 a + 2 a = 8 a \]

Step 7: Use ratio \( r : s = 1 : 16 \)
- Length of latus rectum \( r = 4 a \)
- \( s = 16 a \alpha = 16 a \times 8 a = 128 a^2 \)
- Ratio given:
\[ \frac{r}{s} = \frac{4 a}{128 a^2} = \frac{1}{32 a} = \frac{1}{16} \] - Solve for \( a \):
\[ \frac{1}{32 a} = \frac{1}{16} \implies 32 a = 16 \implies a = \frac{1}{2} \]

Step 8: Find \( 24 a \)
\[ 24 a = 24 \times \frac{1}{2} = 12 \]

Final Answer:
\[ \boxed{12} \]