Let chord PQ of length 3√13 of the parabola y^2=12x be such that the ordinates of points P…

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 :

Accepted Answer

$\begin{gathered} {y}^{2}=12x,4a=12\Rightarrow a=3. \ {y}^{2}=4ax(a{t}^{2},2at). \ yP1:2Q. \ P(3{t}^{2},6t)Q(3(2t{)}^{2},6(2t))=(12{t}^{2},12t). \ PQ: \ PQ=\sqrt{(12{t}^{2}-3{t}^{2}{)}^{2}+(12t-6t{)}^{2}} \ \, =\sqrt{(9{t}^{2}{)}^{2}+(6t{)}^{2}}=\sqrt{81{t}^{4}+36{t}^{2}}. \ PQ=3\sqrt{13}\Rightarrow P{Q}^{2}=117. \ 81{t}^{4}+36{t}^{2}=117 \ \Rightarrow 9{t}^{4}+4{t}^{2}-13=0 \ \Rightarrow ({t}^{2}-1)(9{t}^{2}+13)=0. \ {t}^{2}\ge 0,9{t}^{2}+13
e 0,{t}^{2}=1\Rightarrow t=1. \ P(3,6),Q(12,12). \ {y}^{2}=12xS(a,0)=(3,0). \ QS=\frac{12-0}{12-3}=\frac{12}{9}=\frac{4}{3}. \ PSP(3,6)S(3,0)x. \ \alpha QS \ QS\frac{4}{3}, \ 	an \alpha =\frac{3}{4}. \ \Rightarrow \sin \alpha =\frac{3}{5}. \end{gathered}$

Solutions · 1