$\text{A normal with slope}\frac{1}{\sqrt{\mathrm{ 6}}}\text{is drawn from the point}(0,-\alpha )\text{to the parabola}{\mathrm{ x}}^2=-4ay,\text{where }a>0.$

$\text{Let}\mathrm{ L }\text{be the line passing through} (0,-\alpha )\text{ and parallel to the directrix of the parabola.}$

$\text{Suppose that}\mathrm{ L}\text{ intersects the parabola at two points}\mathrm{ A}\text{ and}\mathrm{ B}.\text{Let }\mathrm{r }\text{denote the length of the}$

$\text{latus rectum and}\mathrm{ s }\text{denote the square of the length of the line segment}\mathrm{ A}B.\text{If }r:s=1:16,\text{then}$

$\text{the value of }24\alpha \text{ is \_\_\_\_\_}$

The correct answer is 12.

We are given the equation of the parabola:

$x^2=-4ay$

Step 1: Finding the equation of the normal to the parabola
Differentiating the equation of the parabola with respect to $x$ , we get:

$2x=-4a\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=-\frac{x}{2a}$

The slope of the tangent at any point $(x_1,y_1)$ on the parabola is $m_T=-\frac{x_1}{2a}$ .
Thus, the slope of the normal $m$ is given by:

$m=-\frac{1}{m_T}=\frac{2a}{x_1}$

We are given that the slope of the normal is $m=\frac{1}{\sqrt{6}}$ . So,

$x_1=2a\sqrt{6}$

Since $(x_1,y_1)$ lies on the parabola $x^2=-4ay$ :

$y_1=-\frac{x_1^2}{4a}=-\frac{24a^2}{4a}=-6a$

The equation of the normal passing through $(2a\sqrt{6},-6a)$ with slope $\frac{1}{\sqrt{6}}$ is:

$y-(-6a)=\frac{1}{\sqrt{6}}(x-2a\sqrt{6})$

$y+6a=\frac{x}{\sqrt{6}}-2a$

Since this normal line passes through the point $(0,-\alpha )$ :

$-\alpha +6a=0-2a\Rightarrow \alpha =8a$

Step 2: Finding the length of segment AB and ratio evaluation
The directrix of the parabola $x^2=-4ay$ is $y=a$ .
The line $L$ passes through $(0,-\alpha )$ and is parallel to the directrix, so its equation is $y=-\alpha$ .
Substituting $y=-\alpha$ into the parabola equation gives the x-coordinates of points $A$ and $B$ :

$x^2=-4a(-\alpha )=4a\alpha \Rightarrow x=\pm 2\sqrt{a\alpha }$

The length of the line segment $AB$ is $4\sqrt{a\alpha }$ , and the square of its length $s$ is:

$s=(4\sqrt{a\alpha }{)}^2=16a\alpha$

The length of the latus rectum $r$ is:

$r=4a$

Using the given ratio $r:s=1:16$ :

$\frac{r}{s}=\frac{4a}{16a\alpha }=\frac{1}{4\alpha }$

Equating to $\frac{1}{16}$ gives:

$4\alpha =2\Rightarrow \alpha =\frac{1}{2}$

Step 3: Calculating 24α
Finally, we calculate the required value:

$24\alpha =24\times \frac{1}{2}=12$