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Q.The equation of the hyperbola with centre at origin, transverse axis along x-axis, length of transverse axis 8 and eccentricity √2 is

a
x216−y216=1\dfrac{x^2}{16} - \dfrac{y^2}{16} = 1
b
x216−y232=1\dfrac{x^2}{16} - \dfrac{y^2}{32} = 1
c
x232−y216=1\dfrac{x^2}{32} - \dfrac{y^2}{16} = 1
d
x232−y232=1\dfrac{x^2}{32} - \dfrac{y^2}{32} = 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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