Q57 of 62 Page 11

Vertices (± 5, 0), the transverse axis is of length.


Since the foci are on x-axis, the equation of the hyperbola is of the form

= 1

Given: foci are (± 5, 0), c = 5

Length of transverse axis = 2a = 8 ⇒ a = 4

Therefore b2 = c2 – a2 = 25 – 16 = 9

= 1

9x2 – 16y2 = 144.

More from this chapter

All 62 →
55
Vertices (0, ± 5), foci (0, ± 8)

56
Vertices (0, ± 3), foci (0, ± 5)

58
Foci (0, ± 13), the conjugate is of length 24.

59
Foci (± 3,0), the latus rectum is of length.