Let chord PQ of length 3√13 of the parabola y^2 = 12x be such that the ordinates of points P and Q are in the ratio 1:2. If the chord PQ subtends an…

Let the coordinates of points $P$P and $Q$Q on the parabola $y^2 = 12x$y2=12x be $(x_1, y_1)$(x1​,y1​) and $(x_2, y_2)$(x2​,y2​).

Given that the ordinates are in the ratio $1:2$1:2, we have $y_2 = 2y_1$y2​=2y1​.

Since $P$P and $Q$Q lie on the parabola, their abscissae are $x_1 = \dfrac{y_1^2}{12}$x1​=12y12​​ and $x_2 = \dfrac{y_2^2}{12} = \dfrac{4y_1^2}{12} = \dfrac{y_1^2}{3}$x2​=12y22​​=124y12​​=3y12​​.

The length of the chord $PQ$PQ is $3\sqrt{13}$313​, so $PQ^2 = 117$PQ2=117.

Using the distance formula:

$$(x_2 - x_1)^2 + (y_2 - y_1)^2 = 117$$(x2​−x1​)2+(y2​−y1​)2=117

$$\left(\dfrac{y_1^2}{3} - \dfrac{y_1^2}{12}\right)^2 + (2y_1 - y_1)^2 = 117$$(3y12​​−12y12​​)2+(2y1​−y1​)2=117

$$\left(\dfrac{y_1^2}{4}\right)^2 + y_1^2 = 117$$(4y12​​)2+y12​=117

$$\dfrac{y_1^4}{16} + y_1^2 - 117 = 0$$16y14​​+y12​−117=0

$$y_1^4 + 16y_1^2 - 1872 = 0$$y14​+16y12​−1872=0

$$(y_1^2 + 52)(y_1^2 - 36) = 0$$(y12​+52)(y12​−36)=0

Since $y_1^2 > 0$y12​>0, we get $y_1^2 = 36$y12​=36, which gives $y_1 = 6$y1​=6 (taking the positive root by symmetry).

Substituting $y_1 = 6$y1​=6, we get $x_1 = \dfrac{36}{12} = 3$x1​=1236​=3. Thus, $P$P is $(3, 6)$(3,6).

For $Q$Q, $y_2 = 12$y2​=12 and $x_2 = \dfrac{144}{12} = 12$x2​=12144​=12. Thus, $Q$Q is $(12, 12)$(12,12).

The focus of the parabola $y^2 = 12x$y2=12x is $S(3, 0)$S(3,0).

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

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

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

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

$$\cos\alpha = \dfrac{\overrightarrow{SP} \cdot \overrightarrow{SQ}}{|\overrightarrow{SP}| |\overrightarrow{SQ}|}$$cosα=∣SP∣∣SQ​∣SP⋅SQ​​

$$\cos\alpha = \dfrac{0(9) + 6(12)}{6 \times \sqrt{9^2 + 12^2}} = \dfrac{72}{6 \times 15} = \dfrac{72}{90} = \dfrac{4}{5}$$cosα=6×92+122​0(9)+6(12)​=6×1572​=9072​=54​

Therefore, $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \left(\dfrac{4}{5}\right)^2} = \dfrac{3}{5}$sinα=1−cos2α​=1−(54​)2​=53​

Answer: $\dfrac{3}{5}$53​