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Chord PQ subtending Right angle at centre of ellipse.

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data

    If P and Q are two points on ellipse [(x^2/a^2)+(y^2/b^2)]=1 such that PQ subtends a right angle at the centre O then.Prove that 1/[(OP)^2] + 1/[(OQ)^2] = [1/(a^2)] +[1/(b^2)]

    2. Relevant equations
    Parametric form of points P(acos(θ),bsin(θ)),


    3. The attempt at a solution
    Since the angle between them is 90 degrees therefore the points are P[acos(θ),bsin(θ)]
    and Q[acos(90+θ),bsin(90+θ)] = Q[-asin(θ),bcos(θ)]

    FirstNow I tried the distance formula but in vain
    SecondI also tried using vectors as follows
    Position vector of P= acos(θ) i +bsin(θ) j
    Q= -asin(θ)i +bcos(θ) j

    Since OP and OQ are perpandicular their dot product is zero
    OP.OQ = 0
    acos(θ).[-asin(θ)] + bsin(θ).[bcoz(θ)]=0
    b^2[sin(θ)cos(θ)]- a^2[sin(θ)cos(θ) = 0
    which gives a^2=b^2
    a = b

    I have 2 questions, 1)How to prove this?
    2)Why does the vector method give a=b (which i think is only a special case). Did I do anything wrong?


    Any help will be appreciated...Thanks.
     
    Last edited by a moderator: Apr 4, 2013
  2. jcsd
  3. Apr 3, 2013 #2

    haruspex

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    Theta is only a parameter here. It does not represent the angle the vector makes to the x axis.
     
  4. Apr 3, 2013 #3
    Oh! Interesting...I had failed to notice that and took it as the angle between positive x-axis and the position vector of the point.
    Thanks

    Ok, so how do I prove the property then ?
     
  5. Apr 3, 2013 #4

    haruspex

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    let the points be parameterised by independent angles, then use the perpendicularity of the vectors to find the relationship between the two.
     
  6. Apr 3, 2013 #5
    Thanks again..
    I tried solving it assuming parameters A and B [P(acosA,bsinA)] and [Q(acosB,bsinB)]
    Therefore OP.OQ = a^2(cosA)(cosB) + b^2(sinA)(sinB) = 0

    Now I simplified it in two ways leading to two equations
    1) (b^2+a^2)[cos(A-B)]=(b^2-a^2)[cos(A+B)]

    2) -(a^2)/(b^2) = tanAtanB

    What should I do next?

    Thanks.
     
  7. Apr 4, 2013 #6

    haruspex

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    The LHS of the target equality is 1/[(OP)^2] + 1/[(OQ)^2], so obtain an expression for that in terms of the angles.
     
  8. Apr 4, 2013 #7
    I have been trying to simplify the equation but in vain....It is too messy and long. Please let me know if you have a better way.
    Anyways thanks for assistance. :)
     
  9. Apr 4, 2013 #8

    haruspex

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    Try substituting t = tan A, u = tan B, and converting all the trig functions into references to t and u. You can use tu = -a2/b2 to eliminate one of them.
     
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