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!

Homework Help: Normal to the hyperbola Question

  1. Jul 24, 2007 #1
    1. The problem statement, all variables and given/known data
    The normal to the hyperbola (x^2)/2 - y^2 = 1 at P (sqrt 3, sqrt 0.5) cuts the y-axis at A and the x-axis at B. Show that PA:PB = 2:1

    2. Relevant equations

    Equation of normal to general hyperbola at (x1,y1) is x(a^2)/x1 + y(b^2)/y1 = a^2 + b^2

    3. The attempt at a solution

    Okay so that makes the eqn of this particular normal: 2x/sqrt 3 + (sqrt 2)y = 3
    So A [0, (sqrt 3)/2] and B [3(sqrt 3)/2, 0]. So using the internal division of lines formula given the ratio 2:1 to prove that the point which cuts the line in that particular ratio is P, I end up with:

    x = [1*0 + 2*3(sqrt3)/2] / (1+2)
    = sqrt 3
    y = [1*(sqrt 3)/2 + 2*0] / (1+2)
    = 1/(2sqrt 3)

    Which is not P... any insights into where I went wrong?
  2. jcsd
  3. Jul 24, 2007 #2
    Are you sure your equations are right? Maybe Im wrong, but shouldnt it be a^2-b^2 on the RHS?
  4. Jul 24, 2007 #3
    Thanks for replying! I'm pretty sure it is a^2 + b^2 though, the equation was given in the question before and I quickly derived it to double check.
  5. Jul 25, 2007 #4
    You have made a major algebraic error in finding to coordination of A and B. I notice A, B and P is not colinear
  6. Jul 25, 2007 #5
    Sorry, I'm a bit confused. How come A,B and P aren't collinear? Aren't they on the same line (the normal to the hyperbola?)
  7. Jul 25, 2007 #6
    Yes... I meant in your solution, A [tex] (0,\sqrt{3}/2) [/tex], B [tex] (3\sqrt{3}/2, 0) [/tex], and P[tex] (\sqrt{3}, \sqrt{0.5}) [/tex] is not collinear..
  8. Jul 25, 2007 #7
    Ohh! I see now, A should be (0, 3/sqrt2). I feel so stupid. Thanks for your help!
  9. Jul 25, 2007 #8
    Sorry... confused it with the equation for the ellipse... my mistake
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook