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JEE Main 2005Medium

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

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

Correct Answer: Option A

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

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