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Math Help - 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 \<br />
\frac{{x^2 }}{{a^2 }} - \frac{{y^2 }}{{b^2 }} = 1<br />
\. S is the focus \<br />
(ae,0)<br />
\ and S' is the focus \<br />
( - ae,0)<br />
\, where e is the eccentricity. Show that \<br />
\left| {SP - S'P} \right| = 2a<br />
\

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

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

    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
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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 \<br />
\frac{{x^2 }}{{a^2 }} - \frac{{y^2 }}{{b^2 }} = 1<br />
\. S is the focus \<br />
(ae,0)<br />
\ and S' is the focus \<br />
( - ae,0)<br />
\, where e is the eccentricity. Show that \<br />
\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 \<br />
\left| {SP + S'P} \right| = 2a \ ....

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

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

    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
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    So I should be trying to prove  \<br />
\left| {SP - S'P} \right| = 2ae<br />
\? I still don't get that though, but
    \<br />
b^2  = a^2 (e^2  - 1)<br />
\
    so rearranging
     \<br />
\sqrt {\frac{{b^2 }}{{a^2 }} + 1}  = e<br />
\
    which is a lot closer to my answer, only I've got an x instead of an a.
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