rearranging formula q

**jharrisbr8** · May 2nd 2011, 11:50 AM

Hi everybody, can somebody tell me were I'm going wrong, this is driving me mad

Make p the subject

$\frac{{x}^{2 } }{{p}^{2 } } + \frac{{y}^{ 2} }{{q}^{ 2} }$ =1

I get

$\frac{x}{\sqrt[]{{q}^{2 }-{y}^{2 }}}$ =p

Which apparently is incorrect. I'd love to show you the steps I took, but this is my first ever post and I'm really struggling with the latex workaround.

**SpringFan25** · May 2nd 2011, 12:01 PM

Im wont try and diagnose your mistake without seeing your working. My attempt:

**jharrisbr8** · May 2nd 2011, 12:54 PM

Hmmm, I get what you've done, but the answer the book gives is

$p=\frac{xq}{\sqrt[]{{q}^{2 }-{y}^{2 }}}$

**TheEmptySet** · May 2nd 2011, 01:32 PM

Originally Posted by jharrisbr8

Hmmm, I get what you've done, but the answer the book gives is

$p=\frac{xq}{\sqrt[]{{q}^{2 }-{y}^{2 }}}$

The answers are equivalent if

$q > 0$

$p=\frac{x}{\sqrt{1-\frac{y^2}{q^2}}}=\frac{x}{\sqrt{\frac{q^2-y^2}{q^2}}} =\frac{x}{\frac{\sqrt{q^2-y^2}}{|q|}}=\frac{|q|x}{\sqrt{q^2-y^2}}$

**jharrisbr8** · May 2nd 2011, 01:47 PM

Got it!

What a relief, thanks Guys.

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