A normal with slope is drawn from the point to the parabola , where . Let be the line passing through and parallel to the directrix of the parabola. Suppose that intersects the parabola at two points A and B. Let denote the length of the latus rectu

Question

A normal with slope $\frac{1}{\sqrt{6}}$ is drawn from the point $(0, - α)$ to the parabola $x^{2} = - 4 a y$ , 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 $24 a$ is ________ . [2024]

Smart Achievers · Accepted Answer

(12)

$x^{2} = - 4 a y$

$Equation of normal$

$y = m x - 2 a - \frac{a}{m^{2}}$

$- α = - 2 a - \frac{a}{\frac{1}{6}} = - 8 a$

$\Rightarrow α = 8 a$

Equation of required line passing through $(0, - α)$ and parallel to directrix is

$y = - α \Rightarrow y = - 8 a,$ Solving with $x^{2} = - 4 a y$

$x^{2} = 32 a^{2} \Rightarrow x = \pm 4 \sqrt{2} a = \pm \frac{α}{\sqrt{2}}$

$A (\frac{α}{\sqrt{2}}, - α), B (- \frac{α}{\sqrt{2}}, - α) \Rightarrow A B = \sqrt{2} α$

$\Rightarrow \frac{r}{s} = \frac{4 a}{2 α^{2}} = \frac{1}{16} \Rightarrow \frac{4 a}{2 \times 64 a^{2}} = \frac{1}{16}$

$\Rightarrow a = \frac{1}{2} \Rightarrow 24 a = 12$

Solution

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Solution

Ans. (12) x2=-4ay Equation of normal y=mx-2a-am2 -α=-2a-a16=-8a ⇒α=8a Equation of required line passing through (0,-α) and parallel to directrix is y=-α⇒y=-8a, Solving with x2=-4ay x2=32a2⇒x=±42a=±α2 A(α2,-α), B(-α2,-α)⇒AB=2α ⇒rs=4a2α2=116⇒4a2×64a2=116 ⇒a=12⇒24a=12

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