Given \(x^2 + 2y^2 - 2x - 4y = -1\)

\((x^2 - 2x) + 2 (y^2 - 2y) = -1\)

\((x^2 - 2x + 1) + 2 (y^2 - 2y + 1) = -1 + 1 + 2\)

\((x - 1)^2 + 2(y - 1)^2 = 2\)

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

\(\frac{(x - 1)^2}{2} + (y - 1)^2 = 1\) It is in the form of \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\)

\(a^2 = 2 , b^2 = 1\)

\(a = \sqrt2 , b = 1\) Distance between center and focus \(c = \sqrt{2 - 1}\)

\(= 1\) Hence Center\((h, k) = (1, 1)\) Foci \(= (h \pm c, k)\)

\(=(1 \pm 1, 1)\) Vertices \(= (h \pm a, k)\)

\(=(1 \pm \sqrt2, 1)\)

\((x^2 - 2x) + 2 (y^2 - 2y) = -1\)

\((x^2 - 2x + 1) + 2 (y^2 - 2y + 1) = -1 + 1 + 2\)

\((x - 1)^2 + 2(y - 1)^2 = 2\)

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

\(\frac{(x - 1)^2}{2} + (y - 1)^2 = 1\) It is in the form of \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\)

\(a^2 = 2 , b^2 = 1\)

\(a = \sqrt2 , b = 1\) Distance between center and focus \(c = \sqrt{2 - 1}\)

\(= 1\) Hence Center\((h, k) = (1, 1)\) Foci \(= (h \pm c, k)\)

\(=(1 \pm 1, 1)\) Vertices \(= (h \pm a, k)\)

\(=(1 \pm \sqrt2, 1)\)