Engineering Board Exam Reviewer · Hyperbolas · easy

Find the foci of the hyperbola x2/16-y2/9=1.

  1. ( 4,0)
  2. ( 5,0)
  3. ( 7,0)
  4. (0, 5)

Answer

B. ( 5,0)

Solution

For a hyperbola c2=a2+b2=16+9=25 c=5. The transverse axis is horizontal, so the foci are ( 5,0).

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  1. Find the length of the transverse axis of the hyperbola x2/16-y2/9=1.
  2. Find the distance between the foci of the hyperbola 9x2-16y2-36x-32y-124=0.
  3. Find the distance between the two directrices of the hyperbola x2/16-y2/9=1.
  4. The point P lies on the rectangular hyperbola xy=18 and has x=3. Find its distance from the origin.
  5. For any point on the hyperbola x2/25-y2/144=1, find the absolute difference of its distances from the two foci.
  6. Find the vertices of the hyperbola (x-1)2/9-(y+2)2/4=1.
  7. Find the equation of the hyperbola with vertices at (0, 4) and asymptotes y= 2x.
  8. Find the length of the latus rectum of the hyperbola x2/9-y2/16=1.
  9. Find the equation of the hyperbola with foci at ( 5,0) and a transverse axis of length 6.
  10. Find the eccentricity of the hyperbola x2/16-y2/9=1.

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