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