Conic section in standard form

Question

From using eigenvectors, completing the square and substitution of values to $s$ and $t$, I have attained the following from a conic section equation:

$s^2+3t^2-\frac{8}{9}=0$

How do I put this into standard form? Wolfram is saying that it is an ellipse so I'm assuming I haven't made mistakes in getting to this equation, otherwise I will check again if there seems to be no answer.

So far, I have done

$s^2+3t^2=\frac{8}{9}$

$\frac{9s^2}{8}+\frac{27t^2}{8}=1$

To try and get it into an ellipse equation but unsure where to go from here

Answer 1

bring 8/9 to RHS and divide to get standard form as s^2/[√(8/9)]^2 + t^2/[√(8/27)^2] = 1 and Wolfram alpha is correct it's an ellipse.

Answer 2

Note that your formula: $$\frac{9s^2}{8}+\frac{27t^2}{8}=1$$ represents an ellipse with centre $O(0,0)$, because is in the form: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Rember also that squaring both sides is dangerous because you obtain only a part of the ellipse. For example here: $$s=\frac{2\sqrt2}{3}\sqrt{1-\frac{27t^2}{8}}$$ you have only the part of the ellipse such that $s\geq0$.