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: