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

Now the center of the ellipse is not the origin of the plane but any point $C$: $C=(x_0,y_0)$.

In this case we will think that the focal axis is parallel to the x-axis, and therefore the foci are in the coordinates $F' (x_0-c,y_0)$ and $F(x_0+c,y_0)$.

Applying these foci in the general definition of the ellipse $$\overline{PF}+\overline{PF'}=2a$$ we obtain $$\sqrt{(x-x_0+c)^2+(y-y_0)^2}+\sqrt{(x-x_0-c)^2+(y-y_0)^2}=2a$$

Moving one of the square roots to the other side and square it: $$\Big(\sqrt{(x-x_0+c)^2+(y-y_0)^2}\Big)^2=\Big(2a-\sqrt{(x-x_0-c)+(y-y_0)}\Big)^2$$ $$(x-x_0+c)^2+(y-y_0)^2=4a^2-4a\sqrt{(x-x_0-x)^2+(y-y_0)^2}+(x-x_0-c)^2+(y-y_0)^2$$ $$(x-x_0)^2+2(x-x_0)c+c^2+(y-y_0)^2= 4a^2-4a\sqrt{(x-x_0-c)^2+(y-y_0)^2}+$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +(x-x_0)^2-2(x-x_0)c+c^2+(y-y_0)^2$$

Simplifying and dividing by four: $$4(x-x_0)c=4a^2-4a\sqrt{(x-x_0-c)^2+(y-y_0)^2}$$ $$(x-x_0)c=a^2-a\sqrt{(x-x_0-c)^2+(y-y_0)^2}$$

Simplifying further, $$(a^2-c(x-x_0))^2=\Big(a \sqrt{(x-x_0-c)^2+(y-y_0)^2}\Big)^2$$ $$a^4-2a^2c(x-x_0)+c^2(x-x_0)^2= a^2\Big((x-x_0-c)^2+(y-y_0)^2\Big)$$ $$a^4-2a^2c(x-x_0)+c^2(x-x_0)^2= a^2\Big((x-x_0)^2-2c(x-x_0)+c^2+(y-y_0)^2\Big)$$ $$a^4-2a^2c(x-x_0)+c^2(x-x_0)^2=a^2(x-x_0)^2-2a^2c(x-x_0)+a^2c^2+a^2(y-y_0)^2$$ $$c^2(x-x_0)^2-a^2(x-x_0)^2-a^2(y-y_0)^2=a^2c^2-a^4$$ $$(c^2-a^2)(x-x_0)^2-a^2(y-y_0)^2= a^2(c^2-a^2)$$

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

Applying the definition $a^2=b^2+c^2$, we replace $-b^2=c^2-a^2$ and we obtain the equation: $$\displaystyle \frac{(x-x_0)^2}{a^2}-\frac{(y-y_0)^2}{-b^2}= 1 \Longrightarrow \frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$$

Therefore the equation is $$\displaystyle \frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$$ and the corresponding graph is:

Let's find the equation of the ellipse centred in $(4,2)$ and with foci $(7,2)$ with major semiaxis $5$.

To calculate $c$ we only need to subtract the $x$ component of the focus from the $x$ component of the center, that is $c=7-4=3$.

Also we know that $a=5$, so we can also compute $b$ thanks to $a^2=b^2+c^2$ and obtain $$b^2=5^2-3^2=25-9=16$$ $$b=4$$ Therefore, substituting in the equation, we see that the expression of the ellipse in question is: $$\displaystyle \frac{(x-4)^2}{5^2}+\frac{(y-2)^2}{4^2}=1$$