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Get the center and the semimajor/semiminor axes of the following ellipses:

$$x^2-6x+4y^2=16$$

$$2x^2 - 4x+3y^2+6y=7$$

How would one get these? I have no clue. I have a problem with merely rewriting these in the traditional ellipse equation.

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2 Answers 2

up vote 1 down vote accepted

$$x^2-6x+4y^2=16$$

$$(x-3)^2-9+4y^2=16$$

$$(x-3)^2+4y^2=25$$

$$\frac{(x-3)^2}{25}+\frac{4y^2}{25}=1$$

$$\frac{(x-3)^2}{5^2}+\frac{y^2}{(\frac {5}{2})^2}=1$$

center is $O(3,0)$ AND axes are $a=5,b=5/2$, for second ellipse you can proceed similarly

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The second one is really tough, I am stuck at $2(x-1)^2 + 3(y+1)^2 =12$ Am I on the right path or not? –  ZafarS Nov 2 '12 at 23:23
    
yes. then divide by 12 and go further –  Adi Dani Nov 2 '12 at 23:28
    
Ok, I'll continue –  ZafarS Nov 2 '12 at 23:29
    
$ M(1,-1)$ and $a=\sqrt{6}$ and $b=2$? –  ZafarS Nov 2 '12 at 23:31
    
that is correct –  Adi Dani Nov 2 '12 at 23:34

I will do the first one: $$ x^2 - 6 x + 4 y^2 = 16 \Rightarrow \left(x - 3\right)^2 + 4 y^2 = 25 \Rightarrow \frac{\left(x-3\right)^2}{5^2} + \frac{y^2}{\left(5/2\right)^2} = 1 $$

Now compare with equation 12 here.

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