Find the coordinates of foci the eccentricity for the hyperbola 9x2 ndash 16y2 ndash 72x 96y ndash 1...........

Mathematics (NDA) Coordinate Geometry

Find the coordinates of foci, the eccentricity for the hyperbola 9x² – 16y² – 72x + 96y – 144 = 0?

e=5/4 ,focus=(9, 3),(–1, 3)

e=3/5 ,focus=(9, 3),(–1, 3)

e=5/2 ,focus=(2, 3),(–1, 3)

e=5/3 ,focus=(9, 1),(–1, 3)

Equation of hyperbola 9x²-16y2 – 72x + 96y – 144 = 0

(x-4)/4² -(y-3)/3² =1

a=4 ,b=3

b²= a² (e² – 1)

9=16(e² – 1)

25/16 =e²

eccentricity= 5/4

Focus X = ± ae , Y = 0 gives the focus as (9, 3), (–1, 3)

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