Can you make 'P' the subject please?

[Angry Badger]

Hi,

I'm stuck on this! Have had several attempts but can't arrive at the given solution. I've no propensity for maths!

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Any help gratefully appreciated.

 

[MrAl]

Hi,

That's an ellipse.

p^2=x^2/(1-y^2/q^2)

so

p1=sqrt(x^2/(1-y^2/q^2)), y^2/q^2!=1
p2=-sqrt(x^2/(1-y^2/q^2)), y^2/q^2!=1

BTW, p is the distance along the x axis from the center to each edge of the ellipse.

 

[Angry Badger]

Hello Mr Al,

Don't know if your still around. Thanks for your prompt reply earlier.

Is this how you arrived at your solution?

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That was one of the solutions that I arrived at but the solution in the text book is:

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Actually, in the text book solution the x is X (capitalised). Must be a typo. Is the solution correct?

Thanks again.

 

[MrAl]

Hi again,


Oh ok, well that requires one little extra step then...

starting again with:

p^2=x^2/(1-y^2/q^2)

multiply the top and bottom of the right side by q^2 to get:

p^2=x^2*q^2/(q^2-y^2)

Now when we take the square root we can perform this
on the top but not on the bottom so we get:

p=x*q/sqrt(q^2-y^2)

Ok now?

 

[Angry Badger]

Hi,

Thanks. I can follow your working, I'd never have 'seen' the way to proceed.