Q. 15.0( 1 Vote )

Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is
A. 16x2 – 9y2 = 144

B. 9x2 – 16y2 = 144

C. 25x2 – 9y2 = 225

D. 9x2 – 25y2 = 81

Answer :

Given: Vertices are (±3, 0) and foci are (±5, 0)


To find: equation of the hyperbola


Formula used:


Standard form of the equation of hyperbola is,



Vertices of hyperbola are given by (±a, 0)


Foci of hyperbola are given by (±ae, 0)


Vertices are (±3, 0) and foci are (±5, 0)


Therefore,


a = 3 and ae = 5


3 × e = 5



b2 = a2(e2 – 1)







b2 = 16


Equation of hyperbola:





16x2 – 9y2 = 144


Hence, required equation of hyperbola is 16x2 – 9y2 = 144

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