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Will the real TeeEm please stand up! watch

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    (Original post by TeeEm)


    solution to follow ... hope I have not got my usual typos ...
    Zacken you have the time it will take me to write a neat solution to bAeat me !!
    Argh, this is such a distraction. I was trying to do FP3. I'll give this a go now! How'd your PDE go by the way?
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    (Original post by Zacken)
    Argh, this is such a distraction. I was trying to do FP3. I'll give this a go now! How'd your PDE go by the way?
    Done !!
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    Do you write any post-graduate problems for us to have a go at?
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    (Original post by hezzlington)
    Do you write any post-graduate problems for us to have a go at?
    I am not that good ...
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    (Original post by TeeEm)
    Done !!
    Spoiler:
    Show
    \dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1

    When x=-ae

    y^{2}=b^{2}(1-e^{2})

    y=b\sqrt{1-e^{2}}

    tan\theta = \dfrac{b\sqrt{1-e^{2}}}{2ae}

    b=a\sqrt{1-e^{2}}\ \ \rightarrow \ \  tan\theta = \dfrac{a\sqrt{1-e^{2}}\sqrt{1-e^{2}}}{2ae}

    tan\theta = \dfrac{1-e^{2}}{2e}
    This along the right lines?
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    (Original post by Zacken)
    Argh, this is such a distraction. I was trying to do FP3. I'll give this a go now! How'd your PDE go by the way?
    you have around 10 minutes ...

    Name:  b8.jpg
Views: 68
Size:  5.8 KB

    I am scanning today's work and I will post straight after.
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    (Original post by edothero)
    Spoiler:
    Show
    \dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1

    When x=-ae

    y=b\sqrt{1-e^{2}}

    tan\theta = \dfrac{b\sqrt{1-e^{2}}}{2ae}

    b=a\sqrt{1-e^{2}}\ \ \rightarrow \ \  tan\theta = \dfrac{a\sqrt{1-e^{2}}\sqrt{1-e^{2}}}{2ae}

    tan\theta = \dfrac{1-e^{2}}{2e}
    This along the right lines?
    This is not what I had in mind ...
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    (Original post by TeeEm)
    This is not what I had in mind ...
    Well...


    :vroam:
    • Thread Starter
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    (Original post by edothero)
    Well...


    :vroam:
    It is on its way ... I have 30 pages to scan first ..
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    (Original post by TeeEm)
    you have around 10 minutes ...

    Name:  b8.jpg
Views: 68
Size:  5.8 KB

    I am scanning today's work and I will post straight after.
    Okay, go for it. I think I've got it... I used a quadratic solving method though, you won't be happy.
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    (Original post by Zacken)
    Okay, go for it. I think I've got it... I used a quadratic solving method though, you won't be happy.
    it is on its way ...
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    (Original post by TeeEm)
    it is on its way ...
    Excited to see this!
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    (Original post by Zacken)
    Excited to see this!
    I will give you a clue because I am still; scanning my work

    Right angle trigonometry
    Loci property of the ellipse
    then I made it a bit harder with trig identities
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    (Original post by TeeEm)
    I will give you a clue because I am still; scanning my work

    Right angle trigonometry
    Loci property of the ellipse
    then I made it a bit harder with trig identities
    It's the middle bit that I can't quite wrap my head around. Take your time, I can wait. :yep:
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    Zacken
    edothero


    lucky I do not write the FP3 ,,,
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    (Original post by TeeEm)
    Zacken
    edothero


    lucky I do not write the FP3 ,,,
    Aaaaargh! I could have done that, I'm so dumb!
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    (Original post by Zacken)
    Aaaaargh! I could have done that, I'm so dumb!
    you ought to know this ...

    the original problem, with the angle given is 4 lines using this approach.
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    (Original post by TeeEm)
    you ought to know this ...

    the original problem, with the angle given is 4 lines using this approach.
    Yeah, I dooo. I literally just covered it two days ago. I couldn't see a nice way to express the hypotenuse :facepalm:
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    (Original post by TeeEm)
    Zacken
    edothero


    lucky I do not write the FP3 ,,,
    is |SP| + |TP| = 2a a loci property of an ellipse?

    Was never taught that

    Though to be fair it is quite obvious..
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    (Original post by Zacken)
    Yeah, I dooo. I literally just covered it two days ago. I couldn't see a nice way to express the hypotenuse :facepalm:
    it happens ...
    well you learned a bit and I have another nice question to add to my books
 
 
 
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