Parabola and Normals
DAILY_CHALLENGE
Grade None
Question:
A normal with slope $\dfrac{1}{\sqrt{6}}$ 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 ___.
Step-by-Step Solution
Key Concept: Using the slope form of the normal to a parabola to relate the intercepts, and using intersection coordinates to find the length of a chord.
For the parabola $x^2 = -4ay$:
- Length of latus rectum: $r = 4a$.
- Parametric point on parabola: $(2at, -at^2)$.
The slope of the tangent is $-t$, so the slope of the normal is:
$$m_{normal} = \dfrac{1}{t}$$
Given the normal slope is $\dfrac{1}{\sqrt{6}} \implies t = \sqrt{6}$.
The equation of this normal is:
$$y - (-6a) = \dfrac{1}{\sqrt{6}} (x - 2a\sqrt{6}) \implies y = \dfrac{x}{\sqrt{6}} - 8a$$
Since this normal passes through $(0, -\alpha)$:
$$-\alpha = -8a \implies \alpha = 8a$$
So $L$ is the horizontal line $y = -8a$.
Find the intersection of $L$ and the parabola:
$$x^2 = -4a(-8a) = 32a^2 \implies x = \pm 4\sqrt{2}a$$
Thus, $AB = 8\sqrt{2}a$, and its square is:
$$s = (AB)^2 = 128a^2$$
Using the given ratio $r : s = 1 : 16$:
$$\dfrac{4a}{128a^2} = \dfrac{1}{16} \implies \dfrac{1}{32a} = \dfrac{1}{16} \implies a = \dfrac{1}{2}$$
Hence:
$$24a = 24\left(\dfrac{1}{2}\right) = 12$$
Thus, the answer is 12.
Correct Answer: 12