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# algebra question (easy probably)

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ok I'm doing a precalc problem and the book is showing how to go from one equation to another form and I'm just not seeing how he's doing it. this isn't for school (since I don't even go and can't afford it right now) and I'm giving you the answers and want to know HOW he's doing it. so obviously you're not just giving me the answers. after all what's the use of reading if I'm not going to learn it. anyway, here's the equations:
(1)
SQRT( (x+c)^2 + y^2 ) + SQRT( (x-c)^2 + y^2 ) = 2a
(2)
a^2 - cx = a * SQRT( (x-c)^2 + y^2 )
(3)
(a^2 - c^2)x^2 + a^2*y^2 = a^2*(a^2 - c^2)

now I've only attempted going from 1 to 2 so far so I haven't even looked from 2 to 3 but if you want to comment on that also that's fine. also, he says in going from 1 to 2 to that he transposes the 2nd radical (first equation), squares, then reduces to become (2). if that helps. hopefully I typed those out the correct way and there wasn't an easier to read way. thanks. edit: er and i should say i'm not completely dumb here. i've transposed and squared the whole equation like he says but it doesn't turn out right at all. i have no idea how he got that 'a' on the right side of equation 2. [edited by - sporff on June 1, 2004 5:52:24 PM]

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Hi Sporff,

THANK YOU for following the forum rules for "homework-like" posts.

I do advise forum members to only post suggestions/hints rather than the actual steps to derive the answer. I''d like you to use these hints to discover the solution on your own, .

Graham Rhodes
Principal Scientist
Applied Research Associates, Inc.

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well, a hint then.
you can get a to the right side of the equation just by multiplying both sides by a. notice that a is squared on the left

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quote:
Original post by grhodes_at_work
Hi Sporff,
THANK YOU for following the forum rules for "homework-like" posts.
I do advise forum members to only post suggestions/hints rather than the actual steps to derive the answer. I'd like you to use these hints to discover the solution on your own, .

that works fine. I just want to understand how he did it. hints would be great because it'd still let me solve it. but as long as they're enough information for me to actually solve it of course. this really isn't even the problem... this is just deriving something else. I could skip this if I wanted to but I'm really trying to work on my algebra.
but anyway, hints or anything are welcomed. thanks.

quote:
Original post by aaroncox1234
well, a hint then.
you can get a to the right side of the equation just by multiplying both sides by a. notice that a is squared on the left

won't that make me multiply the left side's radical also though? i attempted that but i was unsure of how to get rid of that since an 'a' multiplied by an x's c's or y's is not in equation 2 in his work. just an a^2 - cx.

[edited by - sporff on June 1, 2004 6:11:30 PM]

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Please show us your work to try and simplify equation 1 into equation 2. It doesn''t look right to me either. Show us where *you* believe this breaks down.

Graham Rhodes
Principal Scientist
Applied Research Associates, Inc.

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More specifically, I believe this problem relates to geometry, and we need to know what you know about the geoemtry before we can give you hints as to how the simplification can be done as shown. What have you reasoned about the problem based on what you know beyond these equations?

Graham Rhodes
Principal Scientist
Applied Research Associates, Inc.

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Rearrange like this before squaring both sides:

SQRT( (x+c)^2 + y^2 ) = 2a - SQRT( (x-c)^2 + y^2 )

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this is ugly but i think it should help:

SQRT( (x+c)^2 + y^2 ) + SQRT( (x-c)^2 + y^2 ) = 2a

...square everything...

(x+c)^2 + (x-c)^2 + 2y^2 + 2*SQRT( (x+c)^2 + y^2 )*SQRT( (x-c)^2 + y^2 ) = 2a

... from the original equation:
SQRT( (x+c)^2 + y^2 ) = 2a - SQRT( (x-c)^2 + y^2 )

... so sub 2a - SQRT( (x-c)^2 + y^2 ) in for SQRT( (x+c)^2 + y^2 ) ...

(x+c)^2 + (x-c)^2 + 2y^2 + 2*(2a - SQRT( (x-c)^2 + y^2 ))*SQRT( (x-c)^2 + y^2 ) = 2a

...simplify...

(x+c)^2 + (x-c)^2 + 4a*SQRT( (x-c)^2 + y^2 ) - 2(x-c^2) = 4a^2

sorry its messy, but the above should simplify right. if not, let me know

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the problem is in relation to ellipses in my precalculus book. he takes the first equation which is saying the distance of the 2 focus points of the ellipse (SQRT( (x+c)^2 + y^2 ) + SQRT( (x-c)^2 + y^2 )) is equal to 2a. i'm guessing 'a' is the average distance of each line because if you take a point that is equal distance from the focus points, you will get just a+a.

my problem is he doesn't give any reasoning in the changing of forms of the equation so i have no idea how or why he is doing it. i just know he's taking steps to change equation 1 into the equation for an ellipse. to me it sounded like pure algebra he was using because he does not give any reference to any other reason to why he does it other than saying when you get equation 1 (which i fully understand how to get), we have which, by transposing the 2nd radical (although the first could have been used with same results), squaring, and reducing, becomes (2).

now after equation 3, what he does is takes (a^2 - c^2) in equation 3 and treats it as a separate variable 'b'. then has b^2x^2 + a^2y^2 = a^2b^2 which easily changes into the equation for an ellipse (which i don't think i need to type here since that's not relavent).

[edited by - sporff on June 1, 2004 6:31:09 PM]

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my post definately works, at least on paper. let me know if i made a type-o. heres the end...

(x+c)^2 + (x-c)^2 + 4a*SQRT( (x-c)^2 + y^2 ) - 2(x-c^2) = 4a^2

... multiply out the (x+c)^2 and (x-c)^2 ...

4a*SQRT( (x-c)^2 + y^2 ) = 4a^2 - x^2 - 2xc - x^2 - 2xc - c^2 + x^2 - 2cx + c^2

this simplifies to...

a*SQRT( (x-c)^2 + y^2 ) = a^2 - cx

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