Please can someone check my working?

**free_to_fly** · February 13th 2008, 03:03 AM

*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.

**mr fantastic** · February 13th 2008, 03:23 AM

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.

**free_to_fly** · February 13th 2008, 05:05 AM

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.

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