5. The equation of the hyperbola whose foci are
(-2,0) and (2,0) and eccentricity is 2 is given
[2011]
(a) - 3x2 + y2 = 3 (b)x+ - 3y2 - 3
(c) 3x2 - y2 = 3 (d) - + 3y = 3
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Answer:
y^2 - 5xy^2 + 3x^2 +3
= (3)^2 - 5(3)(3)^2 + 3(3)^2 + 3
= 9 - (15)(9) + (3)(9) + 3
= 9 - 135 + 27 + 3
= 39 - 135
= – 96
Step-by-step explanation:
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