Equation of the ellipse with foci on the axis OY

Now, we will suppose that the foci $$F$$ and $$F'$$ are on the axis $$OY$$, so that they are defined by $$F'=(0,-c)$$ and $$F=(0,c)$$ and therefore the ellipse is also centred on the origin, but now the major semiaxis is the vertical one and the minor semiaxis is horizontal, precisely the opposite to equation ellipse I.

With the same ellipse definition we will write that any point $$P$$ of the ellipse satisfies: $$$\overline{PF}+ \overline{PF'}=2a$$$ where $$a$$ corresponds to a constant that we can determine as: $$a^2=b^2+c^2$$. Let's see it in the following drawing:

We will now develop $$$\overline{PF}+ \overline{PF'}=2a$$$ which is equivalent to the expression $$$\sqrt{x^2+(y-c)^2}+\sqrt{x^2+(y+c)^2}=2a$$$ We isolate the first root and square everything: $$$\Big(\sqrt{(y-c)^2+x^2}\Big)^2=\Big(2a-\sqrt{(y+c)^2+x^2}\Big)^2 $$$ $$$(y-c)^2+x^2=4a^2-2\cdot 2a\sqrt{(y+c)^2+x^2}+(y+c)^2+x^2$$$ $$$ y^2-2yc+c^2+x^2=4a^2-4a\sqrt{(y-c)^2+x^2}+y^2+2yc+c^2+x^2$$$

Now we isolate on one side of the equation the root that we have left, thereby obtaining: $$$4a\sqrt{(y+c)^2+x^2}=4a^2+4yc$$$ $$$\displaystyle a\sqrt{(y+c)^2+x^2}= \frac{4a^2+4yc}{4}=a^2+cy $$$

We raise both sides of the equality to the square: $$$\Big( a\sqrt{(y+c)^2+x^2}\Big)^2=(a^2+cy)^2$$$ $$$a^2((y+c)^2+x^2)=a^4+2a^2cy+c^2y^2 $$$ $$$ a^2(y^2+2cy+c^2+x^2)=a^4+2a^2cy+c^2y^2$$$ $$$a^2y^2+2a^2cy+a^2cy+a^2c^2+a^2x^2=a^4+2a^2cy+c^2y^2$$$

Remembering that this relation exists $$a^2=b^2+c^2$$, we have: $$$(a^2-c^2)y^2+a^2c^2+a^2x^2=a^4$$$ $$$ b^2y^2+a^2x^2=a^4-a^2c^2=a^2(a^2-c^2)=a^2b^2$$$

Now we divide both sides of the expression by the factor $$a^2b^2$$ and it gives: $$$\displaystyle \frac{b^2y^2+a^2x^2}{a^2b^2}=\frac{a^2b^2}{a^2b^2}$$$ $$$\displaystyle \frac{y^2}{a^2}+\frac{x^2}{b^2}=1 \Rightarrow \frac{x^2}{b^2}+\frac{y^2}{a^2}=1$$$

We already have the equation of the ellipse for this second case. As it is possible to appreciate, the formula is very similar, only now the square of the value of the major semiaxis is not under $$x$$, but under $$y$$. The values of the semiaxes of the denominators have been changed.

Let's see an example: