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Thread: Need some clarification

  1. #1
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    Need some clarification

    AB is a diameter of the ellipse $\displaystyle x^2+9y^2=25$ the eccentric angle at A is $\displaystyle \frac{\pi}{6}$. 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 $\displaystyle -\frac{5\pi}{6}$

    When I did (b) and (c), I used $\displaystyle m_2=-\frac{b^2}{a^2m_1}$ to find the gradient of the tangent and conjugate diameter. $\displaystyle m_1=\tan\frac{\pi}{6}$
    But I found that in the answers the gradients are all $\displaystyle \frac{1}{\sqrt{3}}$ which is the gradient of the diameter AB.
    $\displaystyle m_2=-\frac{\sqrt{3}}{9}$
    I used the equation for tangents $\displaystyle y=mx\pm\sqrt{a^2m^2+b^2}$
    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
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    Hello arze
    Quote Originally Posted by arze View Post
    AB is a diameter of the ellipse $\displaystyle x^2+9y^2=25$ the eccentric angle at A is $\displaystyle \frac{\pi}{6}$. 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 $\displaystyle -\frac{5\pi}{6}$

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

    Here's my working.


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

    Grandad
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