1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parabolas finding points

  1. Jun 2, 2012 #1
    The point P(4t^2, 8t) lies on the parabola C with equation y^2 = 16x. The point P also lies on the rectangular hyperbola H with equation xy = 4

    a) Find the value of t, and hence find the co-ords of P.

    working:
    so x = 4t^2 and y = 8t
    i sub these into xy = 4 and get t = 1/2 and can then find the points of P and thus answer the question

    however if i sub x = 4t^2 and y = 8t into y^2 = 16x then I get:
    (8t)^2 = 16(4t^2)
    64t^2 = 64t^2
    t = 0?

    Could anyone tell me what im doing wrong when subbing it into y^2 = 16x?
     
  2. jcsd
  3. Jun 2, 2012 #2

    rock.freak667

    User Avatar
    Homework Helper

    You did nothing wrong, you just ended up with 0 = 0 which is true. So you only real solution it t = 1/2.
     
  4. Jun 2, 2012 #3
    I don't understand, shouldnt I get t = 1/2 no matter where I substitute the values into?
     
  5. Jun 2, 2012 #4

    rock.freak667

    User Avatar
    Homework Helper

    Well for the parabola C, the parameter x=4t2, y = 8t is valid. However for the hyperbola, it is not. So you putting it back into the equation for C would just be verifying the parameters are correct.
     
  6. Jun 4, 2012 #5
    When we arrange for t, the answer will be zero.
     
  7. Jun 4, 2012 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The solution to 64t2 = 64t2 is that t can be any real number, not just t=0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook