The equation to the hyperbola having its eccentricity 2 and the distance between its foci is 8 [Karnataka CET 2005]

\[\frac{{{x}^{2}}}{8}-\frac{{{y}^{2}}}{2}=1\]

\[\frac{{{x}^{2}}}{12}-\frac{{{y}^{2}}}{4}=1\]

\[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{12}=1\]

\[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{9}=1\]

The equation to the hyperbola having its eccentricity 2 and the dist

1.	The equation to the hyperbola having its eccentricity 2 and the distance between its foci is 8 [Karnataka CET 2005]
A.	\[\frac{{{x}^{2}}}{12}-\frac{{{y}^{2}}}{4}=1\]
B.	\[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{12}=1\]
C.	\[\frac{{{x}^{2}}}{8}-\frac{{{y}^{2}}}{2}=1\]
D.	\[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{9}=1\]
Answer» C. \[\frac{{{x}^{2}}}{8}-\frac{{{y}^{2}}}{2}=1\]

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The equation to the hyperbola having its eccentricity 2 and the distance between its foci is 8 [Karnataka CET 2005]

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