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

Consider the focus $S(3, 0)$ of the parabola $y^2 = 12x$ (where $a=3$). Let $P(3t_1^2, 6t_1)$ and $Q(3t_2^2, 6t_2)$. Given $y_2 = 2y_1$, we have $t_2 = 2t_1$.
The focal distance of any point $X(x, y)$ on the parabola $y^2 = 4ax$ is given by $SX = a + x$.
Thus, $SP = 3 + 3t_1^2 = 3(1+t_1^2)$ and $SQ = 3 + 3t_2^2 = 3(1+4t_1^2)$.
The length of chord $PQ$ is $3\sqrt{13}$. In $\triangle SPQ$, by the Law of Cosines:
$$PQ^2 = SP^2 + SQ^2 - 2(SP)(SQ)\cos \alpha$$
First, let's find $t_1$ using the chord length formula: $PQ = a|t_2 - t_1|\sqrt{(t_1+t_2)^2 + 4}$.
$$3\sqrt{13} = 3|t_1|\sqrt{9t_1^2 + 4} \implies 13 = t_1^2(9t_1^2 + 4) \implies 9t_1^4 + 4t_1^2 - 13 = 0$$
This factors to $(t_1^2 - 1)(9t_1^2 + 13) = 0$, giving $t_1^2 = 1$.
Now substitute $t_1^2 = 1$ into the expressions for focal distances:
$SP = 3(1+1) = 6$ and $SQ = 3(1+4) = 15$.
Substituting these into the Law of Cosines equation:
$$(3\sqrt{13})^2 = 6^2 + 15^2 - 2(6)(15)\cos \alpha$$
$$117 = 36 + 225 - 180\cos \alpha \implies 180\cos \alpha = 261 - 117 = 144$$
$$\cos \alpha = \frac{144}{180} = \frac{4}{5}$$
Therefore, $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - (4/5)^2} = \frac{3}{5}$.