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