How can I rearrange this formula?

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Homework Statement


Hi!

I'm reading a great book by Paul Nahin, 'Chases & Pursuits', and during one derivation he skips a few steps in rearranging a formula. I'm struggling to see exactly how it was arranged and it's really bugging me.. it can't be too difficult, I think I'm just missing a key step in the process.

If someone would be kind enough as to give me a clue it would be much appreciated!

Thanks in advance,
Gus

Homework Equations


[itex]\frac{\sqrt{(x-a)^{2}+y^{2}}}{\sqrt{(x-b)^{2}+y^{2}}}=k^{2}[/itex]

where [itex]a[/itex], [itex]b[/itex] and [itex]k[/itex] are constants.

The Attempt at a Solution


(or 'What it should be rearranged to')

[itex]\left [ x-\frac{k^{2}b-a}{k^2-1}\right ]^{2}+y^{2}=\left [\frac{k(a-b)}{1-k^{2}}\right ]^{2}[/itex]
 
Last edited:
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Apologies, I wanted to change p and m from the original text to a and b... OP has been updated.
 
poloshermanos said:
...
If someone would be kind enough as to give me a clue it would be much appreciated!

Thanks in advance,
Gus


Homework Equations


[itex]\frac{\sqrt{(x-a)^{2}+y^{2}}}{\sqrt{(x-b)^{2}+y^{2}}}=k^{2}[/itex]

where [itex]a[/itex], [itex]b[/itex] and [itex]k[/itex] are constants.


The Attempt at a Solution


(or 'What it should be rearranged to')

[itex]\left [ x-\frac{k^{2}b-a}{k^2-1}\right ]^{2}+y^{2}=\left [\frac{k(a-b)}{1-k^{2}}\right ]^{2}[/itex]

Start by multiplying both sides by the denominator. Then square both sides. ...