Conic Sections Question 113
Question: If the eccentricity of the hyperbola $ x^{2}-y^{2}{{\sec }^{2}}\alpha =5 $ is $ \sqrt{3} $ times the eccentricity of the ellipse $ x^{2}{{\sec }^{2}}\alpha +y^{2}=25 $ , then a value of $ \alpha $ is
Options:
A) $ \pi /6 $
B) $ \pi /4 $
C) $ \pi /3 $
D) $ \pi /2 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] For the hyperbola $ \frac{x^{2}}{5}-\frac{y^{2}}{5{{\cos }^{2}}\alpha }=1 $ We have $ e_1^{2}=1+\frac{b^{2}}{a^{2}}=1+\frac{5{{\cos }^{2}}\alpha }{5}=1+{{\cos }^{2}}\alpha $ For the ellipse $ \frac{x^{2}}{25{{\cos }^{2}}\alpha }+\frac{y^{2}}{25}=1 $ We have $ e_2^{2}=1-\frac{25{{\cos }^{2}}\alpha }{25}={{\sin }^{2}}\alpha $ Given that $ e_1=\sqrt{3}e_2 $
$ \therefore e_1^{2}=3e_2^{2} $ Or $ 1+{{\cos }^{2}}\alpha =3{{\sin }^{2}}\alpha $ Or $ 2=4{{\sin }^{2}}\alpha $ Or $ \sin \alpha =\frac{1}{\sqrt{2}} $
BETA
