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Math Help - locus and ellipse problems

  1. #1
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    locus and ellipse problems

    Hi,

    You can probably tell by now that me and conic section/locus questions don't get on.

    This time we are considering an ellipse with equation

    \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and focii at S and S'.

    Consider an arbitrary point P on the ellipse, I need to show that the line SP and the perpendicular from the center O to the tangent at P meet at G, and that the locus of G is a circle centre S with radius a.

    Well the gradient of the tangent at a point P(a cos(t), b sin(t)) is \frac{-b \cos t}{a\sin t} so the gradient of the perpendicular to the tangent is \frac{a \sin t}{b \cos t}, this passes through the origin and so the equation for this line is y= \frac{a \sin t}{b \cos t} x

    Now the gradient of the line SP is \frac{\Delta y}{\Delta x} = \frac{b \sin t}{a \cos t + ae}. This passes through the point (-ae, 0) and so the equation of the line SP is

    y=\frac{b \sin t}{a \cos t + ae}(x+ae)

    Presumably I must solve these equations simultaneously and then eliminate t to find the required equation of the circle. Since there is a lone y on each equation I tried setting them equal but got a horrific mess with the algebra so I think there's a trick I might be missing

    Any help on how to solve these equations simultaneously (or even if my method is correct!) would be very much appreciated thank you.
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  2. #2
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    So you have the equations y= \frac{a \sin t}{b \cos t} x and y=\frac{b \sin t}{a \cos t + ae}(x+ae), and you want to eliminate t between them.

    Write the first equation as \tan t = \frac{by}{ax} and think of by and ax as sides in a right-angled triangle. Pythagoras' theorem tells you that \cos t = ax/D and \sin t = by/D, where D = \sqrt{a^2x^2+b^2y^2}. Substitute those values for cos(t) and sin(t) into the second equation, do a bit of cancelletion, use the fact that 1-e^2 = b^2/a^2, and it simplifies to D = -aex+b^2. Now square both sides to get a^2x^2 + b^2y^2 = a^2e^2x^2 - 2b^2aex + b^4, and a bit more simplification reduces that to (x+ae)^2 + y^2 = a^2.
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  3. #3
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    Brilliant! Thanks
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