to the ellipse \[\frac{{{x}^{2}}}{14}+\frac{{{y}^{2}}}{5}=1\] intersects it again at the point \[Q(2\theta )\], then \[\cos \,\theta \] is equal to
A) \[\frac{2}{3}\]
B) \[-\frac{2}{3}\]
C) \[\frac{3}{2}\]
D) \[-\frac{3}{2}\]
Correct Answer: B
Solution :
The normal to the ellipse at \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[P(a\,cos\theta ,\,\,b\,\,\sin \,\,\theta )\] is, \[\operatorname{ax}\,sec \theta -by\,\,cosec \theta =\,\,{{a}^{2}}- {{b}^{2}}\] It meets the curve again at 0(29) i.e., \[(a\,cos 2\theta ,\,\,b\,sin 2\,\theta )\]. \[\therefore \,\,\frac{a}{\cos \theta }a\cos 2\theta -\frac{b}{\sin \theta }\,(b\,sin2\theta )={{a}^{2}}-{{b}^{2}}\] \[\frac{{{x}^{2}}}{14}+\frac{{{y}^{2}}}{5}=1\] Given ellipse is, \[\frac{{{x}^{2}}}{14}+\frac{{{y}^{2}}}{5}=1\] here, \[{{a}^{2}}=14\,\,and\,\,{{b}^{2}}=5\] \[\Rightarrow \,\,\,\frac{14}{\cos \theta }\cos 2\,\theta -\frac{5}{\sin \theta }(\sin 2\theta )=14-5\] \[\Rightarrow \,\,\,18 co{{s}^{2}}\,\theta - 9 cos \theta - 14 = 0\] \[\Rightarrow \,\,\,\left( 6 cos \theta -7 \right) \left( 3 cos \theta +2 \right)= 0\] \[\Rightarrow \,\,\,\cos \theta =-\frac{2}{3}\] \[\left[ \because \,\,\cos \,\theta =\frac{7}{6}\,\,not\,\,possible \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
