The length of the transverse axis of the hyperbola $9 x^{2} - 16 y^{2} - 18 x - 32 y - 151 = 0 i s$

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Solution

The correct option is A
8

Given hyperbola is $\frac{(x - 1)^{2}}{16} - \frac{(y + 1)^{2}}{9} = 1$

Length of the transverse axis is 2a=8.

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Similar questions

The distance between the foci of the hyperbola $9 x^{2} - 16 y^{2} + 18 x + 32 y - 151$ = 0 is

9 x^{2} - 16 y^{2} + 18 x + 32 y - 151 = 0

9 x^{2} - 16 y^{2} + 18 x + 32 y - 151 = 0

9 x^{2} - 16 y^{2} - 18 x + 32 y - 151 = 0

9 x^{2} - 16 y^{2} - 18 x - 32 y - 151 = 0

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