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Math Help - Need some clarification

  1. #1
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    Need some clarification

    AB is a diameter of the ellipse x^2+9y^2=25 the eccentric angle at A is \frac{\pi}{6}. Find
    a) the eccentric angle of B
    b) the equations of the tangent at A and B
    c) the equation of the conjugate diameter

    I found correctly the eccentric angle of B to be -\frac{5\pi}{6}

    When I did (b) and (c), I used m_2=-\frac{b^2}{a^2m_1} to find the gradient of the tangent and conjugate diameter. m_1=\tan\frac{\pi}{6}
    But I found that in the answers the gradients are all \frac{1}{\sqrt{3}} which is the gradient of the diameter AB.
    m_2=-\frac{\sqrt{3}}{9}
    I used the equation for tangents y=mx\pm\sqrt{a^2m^2+b^2}
    Have I misunderstood something?
    Thanks
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  2. #2
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    Well your method is right but i would suggest you another method that you can use the formula of :
    tanθ = m1 + m2
    1 + m1m2
    You will get answer in the form of m. Then you can find the value by elimination method.
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  3. #3
    MHF Contributor
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    Hello arze
    Quote Originally Posted by arze View Post
    AB is a diameter of the ellipse x^2+9y^2=25 the eccentric angle at A is \frac{\pi}{6}. Find
    a) the eccentric angle of B
    b) the equations of the tangent at A and B
    c) the equation of the conjugate diameter

    I found correctly the eccentric angle of B to be -\frac{5\pi}{6}

    When I did (b) and (c), I used m_2=-\frac{b^2}{a^2m_1} to find the gradient of the tangent and conjugate diameter. m_1=\tan\frac{\pi}{6} Grandad says: No. See below*
    But I found that in the answers the gradients are all \frac{1}{\sqrt{3}} which is the gradient of the diameter AB.
    m_2=-\frac{\sqrt{3}}{9}
    I used the equation for tangents y=mx\pm\sqrt{a^2m^2+b^2}
    Have I misunderstood something?
    Thanks
    Are you sure the gradient of the tangent and the conjugate diameter is given as \frac{1}{\sqrt3}? I make it -\frac{1}{\sqrt3}.

    Here's my working.


    The equation of the tangent at (a\cos t, b\sin t) is easily shown to be:
    \frac{x\cos t}{a}+\frac{y\sin t}{b}=1
    and the gradient at (a\cos t, b\sin t) is
    -\frac{b\cot t}{a}
    Here a = 5,\; b = \frac53 and t = \frac{\pi}{6} at A. So the gradient at A is:
    -\frac{\frac53\cot(\pi/6)}{5}=-\frac{1}{\sqrt3}
    and the tangent at A is
    x\sqrt3+3y=10
    (* Incidentally, the gradient of a diameter isn't \tan t. It's \frac{b\tan t}{a}.)

    Grandad
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