A) \[\frac{2}{3}\]
B) \[-\frac{2}{3}\]
C) \[\frac{3}{2}\]
D) \[-\frac{3}{2}\]
Correct Answer: B
Solution :
The normal at \[P(a\cos \theta ,b\sin \theta )\] is \[ax\sec \theta -by\,\,\text{cosec}\theta ={{a}^{2}}-{{b}^{2}}\], where \[{{a}^{2}}=14,\,\,{{b}^{2}}=5\] It meets the curve again at \[Q(2\theta )\] i.e., \[(a\cos 2\theta ,\,b\sin 2\theta )\]. \[\therefore \frac{a}{\cos \theta }a\cos 2\theta -\frac{b}{\sin \theta }(b\sin 2\theta )={{a}^{2}}-{{b}^{2}}\] \[\Rightarrow \frac{14}{\cos \theta }\cos 2\theta -\frac{5}{\sin \theta }(\sin 2\theta )=14-5\] \[\Rightarrow 18{{\cos }^{2}}\theta -9\cos \theta -14=0\] \[\Rightarrow (6\cos \theta -7)(3\cos \theta +2)=0\Rightarrow \cos \theta =-\frac{2}{3}\].
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