Let chord PQ of length 3 13 of the parabola y^2 = 12x be such that...

Let the coordinates of points $P and Q$ on the parabola $y^2 = 12x$ be $(x_1, y_1) and (x_2, y_2)$.

Given that the ordinates are in the ratio $1:2$, we have $y_2 = 2y_1$.

Since $P and Q$ lie on the parabola, their abscissae are $x_1 = \dfrac{y_1^2}{12}$ and $x_2 = \dfrac{y_2^2}{12} = \dfrac{4y_1^2}{12} = \dfrac{y_1^2}{3}$.

The length of the chord $PQ$ is $3\sqrt{13}$, so $PQ^2 = 117$.

Using the distance formula:

$(x_2 - x_1)^2 + (y_2 - y_1)^2 = 117$

\left(\dfrac{y_1^2}{3} - \dfrac{y_1^2}{12}\right)^2 + (2y_1 - y_1)^2 = 117$<br><br>\left(\dfrac{y_1^2}{4}\right)^2 + y_1^2 = 117$

$\dfrac{y_1^4}{16} + y_1^2 - 117 = 0$

$y_1^4 + 16y_1^2 - 1872 = 0$

$(y_1^2 + 52)(y_1^2 - 36) = 0$

Since $y_1^2 > 0$, we get $y_1^2 = 36$, which gives $y_1 = 6$ (taking the positive root by symmetry).

Substituting $y_1 = 6$, we get $x_1 = \dfrac{36}{12} = 3$. Thus, $P$ is $(3, 6)$.

For $Q$, y_2 = 12 and $x_2 = \dfrac{144}{12} = 12$. Thus, $Q$ is $(12, 12)$.

The focus of the parabola $y^2 = 12x$ is $S(3, 0)$.

The vectors from the focus $S$ to points $P and Q$ are:

$\overrightarrow{SP} = (3 - 3)\hat{i} + (6 - 0)\hat{j} = 6\hat{j}$

$\overrightarrow{SQ} = (12 - 3)\hat{i} + (12 - 0)\hat{j} = 9\hat{i} + 12\hat{j}$

The angle $\alpha$ subtended by $PQ$ at the focus is the angle between $\overrightarrow{SP} and \overrightarrow{SQ}$.

$\cos\alpha = \dfrac{\overrightarrow{SP} \cdot \overrightarrow{SQ}}{|\overrightarrow{SP}| |\overrightarrow{SQ}|}$

$\cos\alpha = \dfrac{0(9) + 6(12)}{6 \times \sqrt{9^2 + 12^2}} = \dfrac{72}{6 \times 15} = \dfrac{72}{90} = \dfrac{4}{5}$

Therefore, $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \left(\dfrac{4}{5}\right)^2} = \dfrac{3}{5}$

Answer: $\dfrac{3}{5}$