For the given hyperbola 9x2−16y2+18x+32y−151=0, which of the following options is/are true:
A
Centre of hyperbola is (−1,1).
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B
Length of transverse axis is 8 units.
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C
Length of conjugate axis is 6 units.
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D
Eccentricity of hyperbola is 3.
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Solution
The correct options are A Centre of hyperbola is (−1,1). B Length of transverse axis is 8 units. C Length of conjugate axis is 6 units. 9x2−16y2+18x+32y−151=0 ⇒9(x2+2x+1)−16(y2−2y+1)=144 ⇒(x+1)216−(y−1)29=1 ∴ Centre is (−1,1) ⇒ Length of transverse axis = 2a=2×4=8 units ⇒ Length of conjugate axis =2b=2×3=6 units ⇒e2=1+b2a2=2516 ∴e=54