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Thread: Please can someone check my working?

  1. #1
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    Please can someone check my working?

    *pokes title*

    The question is:
    P is any point on the hyperbola with the equation $\displaystyle \
    \frac{{x^2 }}{{a^2 }} - \frac{{y^2 }}{{b^2 }} = 1
    \$. S is the focus $\displaystyle \
    (ae,0)
    \$ and S' is the focus $\displaystyle \
    ( - ae,0)
    \$, where e is the eccentricity. Show that $\displaystyle \
    \left| {SP - S'P} \right| = 2a
    \$

    My working:
    I've found the length ofSP and S'P using good old Pythagoras

    $\displaystyle \
    SP = \left[ {(ae - x)^2 + (\frac{{ - b}}{a}\sqrt {x^2 - a^2 } )^2 } \right]^{\frac{1}{2}}
    \$, so
    $\displaystyle \
    SP = a - x\sqrt {\frac{{b^2 }}{{a^2 }} + 1}
    \$
    and similarly $\displaystyle \
    S'P = a + x\sqrt {\frac{{b^2 }}{{a^2 }} + 1}
    \$, which gives me
    $\displaystyle \
    \left| {SP - S'P} \right| = 2x\sqrt {\frac{{b^2 }}{{a^2 }} + 1}
    \$

    If one of the signs changes in SP or S'P then I'd get the right answer. Can someone please check if I made a mistake with the signs? Thanks.
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  2. #2
    Flow Master
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    Quote Originally Posted by free_to_fly View Post
    *pokes title*

    The question is:
    P is any point on the hyperbola with the equation $\displaystyle \
    \frac{{x^2 }}{{a^2 }} - \frac{{y^2 }}{{b^2 }} = 1
    \$. S is the focus $\displaystyle \
    (ae,0)
    \$ and S' is the focus $\displaystyle \
    ( - ae,0)
    \$, where e is the eccentricity. Show that $\displaystyle \
    \left| {SP - S'P} \right| = 2a \ $ Mr F says: This is wrong. It should be 2ae. You can see this for yourself by taking P(a, 0) and P'(-a, 0) ...... It's $\displaystyle \
    \left| {SP + S'P} \right| = 2a \ $ ....


    My working:
    I've found the length ofSP and S'P using good old Pythagoras

    $\displaystyle \
    SP = \left[ {(ae - x)^2 + (\frac{{ - b}}{a}\sqrt {x^2 - a^2 } )^2 } \right]^{\frac{1}{2}}
    \$, so
    $\displaystyle \
    SP = a - x\sqrt {\frac{{b^2 }}{{a^2 }} + 1}
    \$
    and similarly $\displaystyle \
    S'P = a + x\sqrt {\frac{{b^2 }}{{a^2 }} + 1}
    \$, which gives me
    $\displaystyle \
    \left| {SP - S'P} \right| = 2x\sqrt {\frac{{b^2 }}{{a^2 }} + 1}
    \$

    If one of the signs changes in SP or S'P then I'd get the right answer. Can someone please check if I made a mistake with the signs? Thanks.
    I don't have time to check right now, but heed my observatiosn above.
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  3. #3
    Member
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    Mar 2007
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    England
    Posts
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    So I should be trying to prove $\displaystyle \
    \left| {SP - S'P} \right| = 2ae
    \$? I still don't get that though, but
    $\displaystyle \
    b^2 = a^2 (e^2 - 1)
    \$
    so rearranging
    $\displaystyle \
    \sqrt {\frac{{b^2 }}{{a^2 }} + 1} = e
    \$
    which is a lot closer to my answer, only I've got an x instead of an a.
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