Question 706005
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*[tex \LARGE (x^2\ -\ 2x\ +\ 1)\ -\ 4(y^2\ -\ 4y\ +\ 4)\ =\ 20\ +\ 1\ -\ 16]


Note the -4 coefficient OUTSIDE the parentheses so the 1st degree y term has to have a coefficient of -4 (-4 times -4 is +16), then in order to complete the square inside the parentheses you need a +4, and +4 times -4 is -16.  What you add to the left must be added to the right, hence the -16 in the RHS.

From there:


*[tex \LARGE \frac{(x\ -\ 1)^2}{5}\ -\ \frac{4(y\ -\ 2)^2}{5}\ =\ 1]


*[tex \LARGE \frac{(x\ -\ 1)^2}{5}\ -\ \frac{(y\ -\ 2)^2}{\frac{5}{4}}\ =\ 1]


So *[tex \LARGE a\ =\ sqrt{5}] and *[tex \LARGE b\ =\ \frac{\sqrt{5}}{2}]


Can you take it from there?


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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