1. ## Please can someone check my working?

*pokes title*

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

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

$\
SP = \left[ {(ae - x)^2 + (\frac{{ - b}}{a}\sqrt {x^2 - a^2 } )^2 } \right]^{\frac{1}{2}}
\$
, so
$\
SP = a - x\sqrt {\frac{{b^2 }}{{a^2 }} + 1}
\$

and similarly $\
S'P = a + x\sqrt {\frac{{b^2 }}{{a^2 }} + 1}
\$
, which gives me
$\
\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.

2. Originally Posted by free_to_fly
*pokes title*

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

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

$\
SP = \left[ {(ae - x)^2 + (\frac{{ - b}}{a}\sqrt {x^2 - a^2 } )^2 } \right]^{\frac{1}{2}}
\$
, so
$\
SP = a - x\sqrt {\frac{{b^2 }}{{a^2 }} + 1}
\$

and similarly $\
S'P = a + x\sqrt {\frac{{b^2 }}{{a^2 }} + 1}
\$
, which gives me
$\
\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.

3. So I should be trying to prove $\
\left| {SP - S'P} \right| = 2ae
\$
? I still don't get that though, but
$\
b^2 = a^2 (e^2 - 1)
\$

so rearranging
$\
\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.