Conic-Sections Question 561
Question: If the eccentricity of the hyperbola $ x^{2}-y^{2}{{\sec }^{2}}\theta =4 $ is $ \sqrt{3} $ times the eccentricity of the ellipse $ x^{2}{{\sec }^{2}}\theta +y^{2}=16. $ then the value of $ \theta $ equals.
Options:
A) $ \frac{\pi }{6} $
B) $ \frac{3\pi }{4} $
C) $ \frac{\pi }{3} $
D) $ \frac{\pi }{2} $
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Answer:
Correct Answer: B
Solution:
[b] Given: $ x^{2}-y^{2}{{\sec }^{2}}\theta =4 $ and $ x^{2}{{\sec }^{2}}\theta +y^{2}=16 $
$ \Rightarrow \frac{x^{2}}{4}-\frac{y^{2}}{4{{\cos }^{2}}\theta }=1 $ And $ \frac{x^{2}}{16{{\cos }^{2}}\theta }+\frac{y^{2}}{16}=1 $ According to problem $ \frac{4+4{{\cos }^{2}}\theta }{4}=3( \frac{16-16{{\cos }^{2}}\theta }{16} ) $
$ \Rightarrow 1+{{\cos }^{2}}\theta =3(1-cos^{2}\theta )\Rightarrow 4cos^{2}\theta =2 $
$ \Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4},\frac{3\pi }{4} $
BETA
