Answer
Type: Ellipse
Vertices: $(\pm\sqrt{2}+1,2)$
Foci: $(0,2)$ and $(2,2)$
Work Step by Step
Rewrite the equation:
$x^2-2x+2y^2-8y+7=0$
$x^2-2x+1+2y^2-8y+8+7=1+8$
$(x-1)^2+2(y^2-4y+4)+7=9$
$(x-1)^2+2(y-2)^2=2$
$\frac{(x-1)^2}{2}+\frac{(y-2)^2}{1}=1$
Recall: The equation $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ for $a\geq b$ describes an ellipse with the vertices $(\pm a+h,k)$ and the foci $(\pm c+h,k)$ where $c^2=a^2-b^2$.
We have
$a^2=2\to a=\sqrt{2}$
$b^2=1\to b=1$
$c=\sqrt{a^2-b^2}=\sqrt{2-1}=\sqrt{1}=1$
$h=1$ and $k=2$
$\pm a+h=\pm \sqrt{2}+1$
$\pm c+h=\pm 1+1\to 0$ or $2$
So, the given equation describes an ellipse whose the vertices are $(\pm\sqrt{2}+1,2)$ and the foci are $(0,2)$ and $(2,2)$.
