| 
 
 
| Question 75052:  Give the vertices of the hyperbola (x-12)^2/16 - (y-14)^2/25 = 1
 I have choices for answers that are:
 a.) V(12+-3,14)
 b.) V(12+-4,14)
 c.) V(12,14+-3)
 d.) V(12,14+-5)
 Answer by stanbon(75887)
      (Show Source): 
You can put this solution on YOUR website! Give the vertices of the hyperbola (x-12)^2/16 - (y-14)^2/25 = 1 ---------------
 The center is at (12,14)
 The semi-major axis is sqrt25=5 and that axis is vertical.
 Vertices on the major axis are (12, 14+5)=(12,19) and (12,14-5)=(12,9)
 -------------
 The semi-minor axis is sqrt16=4 and that axis is horizontal.
 Vertices on the minor axis are (12+4,14)=(16,14) and (12-4,14)=(8,14)
 =============
 Cheers,
 Stan H.
 | 
  
 | 
 |