Conic Sections Question 184
Question: If the normal at the point $ P(\theta ) $ 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
Options:
A) $ \frac{2}{3} $
B) $ -\frac{2}{3} $
C) $ \frac{3}{2} $
D) $ -\frac{3}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
The normal at $ P(a\cos \theta ,b\sin \theta ) $ is $ ax\sec \theta -bycosec\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} $ .
BETA
