Equation of the ellipse with center (x0, y0) and focal axis parallel to y axis

This equation vs focal axis parallel to $x$-axis, is only modified in that $x$ and $y$ have their roles interchanged, and therefore, they will have the coefficients in the denominator interchanged.

Let's see the demonstration:

The focal axis is now parallel to the $y$ axis, and therefore the foci are at points $F'(x_0,y_0-c)$ and $F(x_0,y_0+c)$.

Applying now the general definition we obtain $$\displaystyle \sqrt{(x-x_0)^2+(y-y_0+c)^2}+\sqrt{(x-x_0)^2+(y-y_0-c)^2}=2a$$

We move one of the roots to the other side and we square both sides: $$\displaystyle \Big(\sqrt{(x-x_0)^2+(y-y_0+c)^2}\Big)^2=\Big(2a-\sqrt{(x-x_0)^2+(y-y_0-c)^2}\Big)^2$$ $$(x-x_0)^2+(y-y_0+c)^2=4a^2-4a\sqrt{(x-x_0)^2+(y-y_0-c)^2}+(x-x_0)^2+(y-y_0-c)^2$$ $$(x-x_0)^2+(y-y_0)^2+2(y-y_0)c+c^2= 4a^2-4a \sqrt{(x-x_0)^2+(y-y_0-c)^2}+$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +(x-x_0)^2+(y-y_0)^2-2(y-y_0)c+c^2$$

Simplifying both sides, we obtain: $$4(y-y_0)c=4a^2-4a\sqrt{(x-x_0)^2+(y-y_0-c)^2}$$ $$(y-y_0)c=a^2-a \sqrt{(x-x_0)^2+(y-y_0-c)^2}$$

We clear the square root and we square the whole expression: $$(c(y-y_0)-a^2)^2= \Big(-a \sqrt{(x-x_0)^2+(y-y_0-c)^2}\Big)^2$$ $$c^2(y-y_0)^2-2a^2c(y-y_0)+a^4= a^2((x-x_0)^2+(y-y_0-c)^2)$$ $$c^2(y-y_0)^2-2a^2c(y-y_0)+a^4=a^2((x-x_0)^2+(y-y_0)^2-2c(y-y_0)+c^2)$$ $$c^2(y-y_0)^2-2a^2c(y-y_0)+a^4=a^2(x-x_0)^2+a^2(y-y_0)^2-2a^2c(y-y_0)+a^2c^2$$ $$c^2(y-y_0)^2-a^2(y-y_0)^2-a^2(x-x_0)^2= a^2 c^2-a^4$$ $$(c^2-a^2)(y-y_0)^2-a^2(x-x_0)^2=a^2(c^2-a^2)$$

Then divide by $a^2(c^2-a^2)$ to obtain 1 on the right: $$\displaystyle \frac{(c^2-a^2)(y-y_0)^2}{a^2(c^2-a^2)}-\frac{a^2(x-x_0)^2}{a^2(c^2-a^2)}=1$$ $$\frac{(y-y_0)^2}{a^2}-\frac{(x-x_0)^2}{(c^2-a^2)}=1 $$

By definition we have $a^2= b^2+c^2$, and thus $-b^2=c^2-a^2$ can be replaced and we obtain: $$\displaystyle \frac{(y-y_0)^2}{a^2}- \frac{(x-x_0)^2}{-b^2}= 1 \Longrightarrow \frac{(y-y_0)^2}{a^2}+\frac{(x-x_0)^2}{b^2}=1 $$.

Determine the equation of an ellipse with center at the point $(1,-1)$ and with a focus at the point $(1,2)$. We also know that it goes through the point $(1,4)$.

First, we must think about in which axis the foci of the ellipse are. Since the center is $(1,-1)$ and a focus is in $(1,2)$, we realize that the first component remains at 1, that is to say, the straight line that unites the center with the focus is the straight line $x=1$.

So we already know that the foci are on a straight line parallel to the y-axis $OY$. If the ellipse goes through point $(1,4)$, the distance from this point (which is also the straight line $x=1$ and therefore it is the major axis) to the center is the difference of its components.

That is: $a=4-(-1)=5$.

In the same way we can argue that the value of $c$ which is the distance from the focus to the center is the subtraction of its $y$ component, which is: $c=2-(-1)=3$.

Since we already have the values of $a$ and $c$, thanks to the relation $a^2=b^2+c^2$, we obtain: $b=\sqrt{25-9}=4$.

Substituting in the equation: $$\displaystyle \frac{(y-y_0)^2}{a^2}+\frac{(x-x_0)^2}{b^2}=1 \Longrightarrow \frac{(y+1)^2}{5^2}+\frac{(x-1)^2}{4^2}=1$$