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 $\overset{―}{P F} + \overset{―}{P F^{'}} = 2 a$ we obtain $\sqrt{(x - x_{0} + c)^{2} + (y - y_{0})^{2}} + \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}} = 2 a$

Moving one of the square roots to the other side and square it: $(\sqrt{(x - x_{0} + c)^{2} + (y - y_{0})^{2}})^{2} = (2 a - \sqrt{(x - x_{0} - c) + (y - y_{0})})^{2}$ $(x - x_{0} + c)^{2} + (y - y_{0})^{2} = 4 a^{2} - 4 a \sqrt{(x - x_{0} - x)^{2} + (y - y_{0})^{2}} + (x - x_{0} - c)^{2} + (y - y_{0})^{2}$ $(x - x_{0})^{2} + 2 (x - x_{0}) c + c^{2} + (y - y_{0})^{2} = 4 a^{2} - 4 a \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}} +$ $+ (x - x_{0})^{2} - 2 (x - x_{0}) c + c^{2} + (y - y_{0})^{2}$

Simplifying and dividing by four: $4 (x - x_{0}) c = 4 a^{2} - 4 a \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}}$ $(x - x_{0}) c = a^{2} - a \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}}$

Simplifying further, $(a^{2} - c (x - x_{0}))^{2} = (a \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}})^{2}$ $a^{4} - 2 a^{2} c (x - x_{0}) + c^{2} (x - x_{0})^{2} = a^{2} ((x - x_{0} - c)^{2} + (y - y_{0})^{2})$ $a^{4} - 2 a^{2} c (x - x_{0}) + c^{2} (x - x_{0})^{2} = a^{2} ((x - x_{0})^{2} - 2 c (x - x_{0}) + c^{2} + (y - y_{0})^{2})$ $a^{4} - 2 a^{2} c (x - x_{0}) + c^{2} (x - x_{0})^{2} = a^{2} (x - x_{0})^{2} - 2 a^{2} c (x - x_{0}) + a^{2} c^{2} + a^{2} (y - y_{0})^{2}$ $c^{2} (x - x_{0})^{2} - a^{2} (x - x_{0})^{2} - a^{2} (y - y_{0})^{2} = a^{2} c^{2} - a^{4}$ $(c^{2} - a^{2}) (x - x_{0})^{2} - a^{2} (y - y_{0})^{2} = a^{2} (c^{2} - a^{2})$

We divide by $a^{2} (c^{2} - a^{2})$ to obtain 1 on the right: $\frac{(c^{2} - a^{2}) (x - x_{0})^{2}}{a^{2} (c^{2} - a^{2})} - \frac{a^{2} (y - y_{0})^{2}}{a^{2} (c^{2} - a^{2})} = 1$ $\frac{(x - x_{0})^{2}}{a^{2}} - \frac{(y - y_{0})^{2}}{(c^{2} - a^{2})} = 1$

Applying the definition $a^{2} = b^{2} + c^{2}$ , we replace $- b^{2} = c^{2} - a^{2}$ and we obtain the equation: $\frac{(x - x_{0})^{2}}{a^{2}} - \frac{(y - y_{0})^{2}}{- b^{2}} = 1 ⟹ \frac{(x - x_{0})^{2}}{a^{2}} + \frac{(y - y_{0})^{2}}{b^{2}} = 1$

Therefore the equation is $\frac{(x - x_{0})^{2}}{a^{2}} + \frac{(y - y_{0})^{2}}{b^{2}} = 1$ and the corresponding graph is:

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Example

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: $\frac{(x - 4)^{2}}{5^{2}} + \frac{(y - 2)^{2}}{4^{2}} = 1$