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 :
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Solution
Sol. [Key Concept: Parametrise $P(3t^{2},6t),Q(12t^{2},12t)$ on $y^{2} = 12x$; the ordinate ratio $1:2$ and chord length fix $t$, then $tan\alpha$ from the slopes to the focus gives $sin\alpha$.]
$${P\left( 3t^{2},6t \right),Q\left( 12t^{2},12t \right) }{PQ = \sqrt{81t^{4} + 36t^{2}} }{81t^{4} + 36t^{2} = 117 }{\Rightarrow 9t^{4} + 4t^{2} - 13 = 0 }{\left( t^{2} - 1 \right)\left( 9t^{2} + 13 \right) = 0 \Rightarrow t = 1 }{P(3,6),Q(12,12) }$$slope of $QS = \frac{4}{3}$
$$tan\alpha = \frac{3}{4},sin\alpha = \frac{3}{5} $$
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