Home/Question #3933
JEE Main 2014Medium

Q.The equation of the hyperbola with foci (0,\pm 5) and eccentricity 5/4 is

a
y216−x29=1\dfrac{y^2}{16} - \dfrac{x^2}{9} = 1
b
y29−x216=1\dfrac{y^2}{9} - \dfrac{x^2}{16} = 1
c
x216−y29=1\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1
d
x29−y216=1\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

Unlock Detailed Solution

Register for Free

Already a member? Login here