Question

Let chord $PQ$ of length $3\sqrt{13}$ 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 angle $\alpha$ at the focus of the parabola, then $\sin \alpha$ is equal to:

• A. $\frac{3}{5}$
• B. $\frac{4}{5}$
• C. $\frac{5}{13}$
• D. $\frac{12}{13}$

Hint:

Find the parametric coordinates of points P and Q using the ordinate ratio, then use the chord length to solve for the parameter $t$. Finally, apply the cosine rule or vector dot product in triangle SPQ to find the angle at the focus.

Answer 1

The Correct Option is A

The given parabola is $y^2 = 12x$. Comparing this with the standard equation $y^2 = 4ax$, we find that $4a = 12$, which gives $a = 3$. The focus $S$ of this parabola is at $(a, 0)$, which is $(3, 0)$.
Any point on this parabola can be represented in parametric form as $(at^2, 2at)$. Let the points $P$ and $Q$ be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ respectively. The ordinates of these points are $y_1 = 2at_1$ and $y_2 = 2at_2$.
According to the problem, the ratio of these ordinates is $1:2$. This implies:
$$\frac{2at_1}{2at_2} = \frac{1}{2} \implies t_2 = 2t_1$$
The length of the chord $PQ$ is given as $3\sqrt{13}$. Using the distance formula for two points in parametric form on a parabola:
$$PQ = \sqrt{(at_2^2 - at_1^2)^2 + (2at_2 - 2at_1)^2} = a(t_2 - t_1)\sqrt{(t_1 + t_2)^2 + 4}$$
Substituting $t_2 = 2t_1$ and $a = 3$ into this equation:
$$3(2t_1 - t_1)\sqrt{(t_1 + 2t_1)^2 + 4} = 3\sqrt{13}$$
$$t_1\sqrt{9t_1^2 + 4} = \sqrt{13}$$
Squaring both sides gives $t_1^2(9t_1^2 + 4) = 13$, which leads to the quadratic equation in $t_1^2$: $9(t_1^2)^2 + 4(t_1^2) - 13 = 0$.
Solving this by factoring: $(9t_1^2 + 13)(t_1^2 - 1) = 0$. Since $t_1^2$ must be a positive real number, we have $t_1^2 = 1$, so $t_1 = \pm 1$.
Let $t_1 = 1$. Then $t_2 = 2$. The points are $P(3, 6)$ and $Q(12, 12)$.
We can find the angle $\alpha$ subtended at the focus $S(3, 0)$ using vectors $\vec{SP}$ and $\vec{SQ}$.
$\vec{SP} = (3-3)\hat{i} + (6-0)\hat{j} = 6\hat{j}$ with magnitude $|\vec{SP}| = 6$.
$\vec{SQ} = (12-3)\hat{i} + (12-0)\hat{j} = 9\hat{i} + 12\hat{j}$ with magnitude $|\vec{SQ}| = \sqrt{9^2 + 12^2} = 15$.
Using the dot product formula $\vec{SP} \cdot \vec{SQ} = |\vec{SP}||\vec{SQ}|\cos\alpha$:
$$(0)(9) + (6)(12) = 6 \times 15 \times \cos\alpha \implies 72 = 90\cos\alpha \implies \cos\alpha = \frac{72}{90} = \frac{4}{5}$$
Since $\sin^2\alpha + \cos^2\alpha = 1$, we have:
$$\sin\alpha = \sqrt{1 - (4/5)^2} = \sqrt{1 - 16/25} = \sqrt{9/25} = \frac{3}{5}$$