Engineering Board Exam Reviewer · Hyperbolas · medium

Find the equation of the hyperbola with foci at ( 5,0) and a transverse axis of length 6.

  1. x29 - y216 = 1
  2. x216 - y29 = 1
  3. x29 + y216 = 1
  4. x29 - y225 = 1

Answer

A. x29 - y216 = 1

Solution

2a=6 a=3 and c=5, so b2=c2-a2=16. The equation is x2/9-y2/16=1.

More Hyperbolas questions

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

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