Conic Sections Question 222
Question: A hyperbola having the transverse axis of length $ 2\sin \theta , $ is confocal with the ellipse $ 3x^{2}+4y^{2}=12. $ Then its equation is
Options:
A) $ x^{2}\cos ec^{2}\theta -y^{2}{{\sec }^{2}}\theta =1 $
B) $ x^{2}{{\sec }^{2}}\theta -y^{2}\cos ec^{2}\theta =1 $
C) $ x^{2}{{\sin }^{2}}\theta -y^{2}{{\cos }^{2}}\theta =1 $
D) $ x^{2}{{\cos }^{2}}\theta -y^{2}{{\sin }^{2}}\theta =1 $
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Answer:
Correct Answer: A
Solution:
[a] Equation of the ellipse is $ 3x^{2}+4y^{2}=12 $
$ \Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{3}=1 $
…… (1) Eccentricity $ e_1=\sqrt{1-\frac{3}{4}}=\frac{1}{2} $ So, the foci of ellipse are $ (1,0) $ and $ (-1,0) $ Let the equation of the required hyperbola be $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $
…… (2) Given $ 2a=2\sin \theta \Rightarrow a=\sin \theta $ Since the ellipse (1) and the hyperbola (2) are confocal, so. The foci of hyperbola are $ (1,0) $ and $ (-1,0) $ too. It the eccentricity, of hyperbola be $ e_2 $ then $ ae_2=1\Rightarrow \sin \theta e_2=1\Rightarrow e_2=\cos ec{{\theta }^{2}} $
$ \therefore b^{2}=a^{2}(e^2_2-1)=sin^{2}\theta (cosec^{2}\theta -1)=cos^{2}\theta $
$ \therefore $ Required equation of the hyperbola is $ \frac{x^{2}}{{{\sin }^{2}}\theta }-\frac{y^{2}}{{{\cos }^{2}}\theta }=1\Rightarrow x^{2}\cos ec^{2}\theta -y^{2}{{\sec }^{2}}\theta =1 $
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