# Thread: Find the center the length of the axes and eccentricity of ellipse.

1. ## Find the center the length of the axes and eccentricity of ellipse.

Find the center the length of the axes and eccentricity of ellipse.
$\displaystyle 2x^2+3y^2-4x-12y+13=0$

2. Originally Posted by varunnayudu
Find the center the length of the axes and eccentricity of ellipse.
$\displaystyle 2x^2+3y^2-4x-12y+13=0$

$\displaystyle 2x^2+3y^2-4x-12y+13= 0$

$\displaystyle 2(x^2 - 2x + 1)+3(y^2-4y+4) - 1 = 0$

$\displaystyle 2(x-1)^2+3(y-2)^2 = 1$

$\displaystyle \dfrac{(x-1)^2}{\frac12}+\dfrac{(y-2)^2}{\frac13} = 1$

Now its in standard form, can you continue?

3. ## Ellipse

Hi -

Originally Posted by varunnayudu
Find the center the length of the axes and eccentricity of ellipse.
$\displaystyle 2x^2+3y^2-4x-12y+13=0$
Complete the square for x:

$\displaystyle 2x^2 - 4x = 2(x^2 - 2x)$

$\displaystyle =2((x-1)^2-1)$

$\displaystyle =2(x-1)^2-2$

and for y:

$\displaystyle 3y^2-12y =3(y^2-4y)$

$\displaystyle =3((y-2)^2-4)$

$\displaystyle =3(y-2)^2-12$

So the equation can be re-written:

$\displaystyle 2(x-1)^2-2+3(y-2)^2-12+13=0$

i.e. $\displaystyle 2(x-1)^2 + 3(y-2)^2=1$

Substitute $\displaystyle u=x-1$ and $\displaystyle v=y-2$:

$\displaystyle 2u^2+3v^2=1$

Re-arrange:

$\displaystyle \frac{u^2}{(\frac{1}{\sqrt{2}})^2} + \frac{v^2}{(\frac{1}{\sqrt{3}})^2}=1$

OK from here?