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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$$x^2-6x+4y^2=16$$
$$(x-3)^2-9+4y^2=16$$
$$(x-3)^2+4y^2=25$$
$$\frac{(x-3)^2}{25}+\frac{4y^2}{25}=1$$
$$\frac{(x-3)^2}{5^2}+\frac{y^2}{(\frac {5}{2})^2}=1$$
center is $O(3,0)$ AND axes are $a=5,b=5/2$, for second ellipse you can proceed similarly
Adi Dani
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1The 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? – JohnPhteven Nov 02 '12 at 23:23
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yes. then divide by 12 and go further – Adi Dani Nov 02 '12 at 23:28
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Ok, I'll continue – JohnPhteven Nov 02 '12 at 23:29
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$ M(1,-1)$ and $a=\sqrt{6}$ and $b=2$? – JohnPhteven Nov 02 '12 at 23:31
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that is correct – Adi Dani Nov 02 '12 at 23:34
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@ZafarS.you are WELCOME – Adi Dani Nov 02 '12 at 23:41
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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.
Eric Angle
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