Let chord PQ of length 3sqrt{13} of the parab | vedclass

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(A) For the parabola $y^2 = 4ax$, we have $4a = 12$, so $a = 3$. Let the coordinates of $P$ and $Q$ be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ respect

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$3$/$5$

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(A) For the parabola $y^2 = 4ax$, we have $4a = 12$, so $a = 3$. Let the coordinates of $P$ and $Q$ be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ respectively.The ratio of ordinates is $2at_1 : 2at_2 = 1 : 2$, which implies $t_2 = 2t_1$.The length of the chord $PQ$ is given by $a(t_2 - t_1) \sqrt{(t_2 + t_1)^2 + 4}$.Substituting $a = 3$ and $t_2 = 2t_1$, we get $3(t_1) \sqrt{(3t_1)^2 + 4} = 3\sqrt{13}$.Thus, $t_1 \sqrt{9t_1^2 + 4} = \sqrt{13}$. Squaring both sides, $t_1^2(9t_1^2 + 4) = 13$. Let $u = t_1^2$, then $9u^2 + 4u - 13 = 0$.Solving for $u$, $(9u + 13)(u - 1) = 0$. Since $u > 0$, $u = 1$, so $t_1 = 1$ and $t_2 = 2$.The points are $P(3, 6)$ and $Q(12, 12)$. The focus $S$ is $(a, 0) = (3, 0)$.The slope of $SP$ is $m_1 = (6 - 0) / (3 - 3) = \infty$ (a vertical line $x = 3$).The slope of $SQ$ is $m_2 = (12 - 0) / (12 - 3) = 12 / 9 = 4/3$.The angle $\alpha$ between the lines is given by $	an \alpha = |(m_2 - m_1) / (1 + m_1 m_2)|$. Since one line is vertical, $	an \alpha = |1 / m_2| = 3/4$ is incorrect; rather, the angle between a vertical line and a line with slope $m$ is $\alpha = |90^\circ - 	heta|$, where $	an 	heta = 4/3$. Thus $	an \alpha = \cot 	heta = 3/4$ is wrong; the angle $\alpha$ is $90^\circ - 	heta$ where $	an 	heta = 4/3$. Therefore $\sin \alpha = \cos 	heta = 3/5$ is also not correct. Let's re-evaluate: The angle $\alpha$ is the angle between the vectors $\vec{SP} = (0, 6)$ and $\vec{SQ} = (9, 12)$.$\cos \alpha = (\vec{SP} \cdot \vec{SQ}) / (|SP| |SQ|) = (0 \cdot 9 + 6 \cdot 12) / (6 \cdot \sqrt{9^2 + 12^2}) = 72 / (6 \cdot 15) = 72 / 90 = 4/5$.Since $\cos \alpha = 4/5$, then $\sin \alpha = 3/5$.

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:

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