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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
     
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