If e1 and e2 are the eccentricities of a hyperbola 3x2 - 3y2 = 25 and its conjugate, then

1) e21 + e22 = 2

2) e21 + e22 = 4

3) e1 + e2 = 4

4) e1 + e2 = √2

Solution: (2) e21 + e22 = 4

Given equation can be rewritten as x2 – y2 = 25 / 3

Here, a2 = 1, b2 = 1

∴e1 = 1 + √b2 / a2 = √1 + 1 = √2

The equation of conjugate hyperbola is – x2 + y2 = 25 / 3

∴e2 = √1 + a2 / b2 = √1 + 1 = √2

∴e21 + e22 = (√2)2 + (√2)2 = 4

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