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

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

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

Simplifying both sides, we obtain: $4 (y - y_{0}) c = 4 a^{2} - 4 a \sqrt{(x - x_{0})^{2} + (y - y_{0} - c)^{2}}$ $(y - y_{0}) c = a^{2} - a \sqrt{(x - x_{0})^{2} + (y - y_{0} - c)^{2}}$

We clear the square root and we square the whole expression: $(c (y - y_{0}) - a^{2})^{2} = (- a \sqrt{(x - x_{0})^{2} + (y - y_{0} - c)^{2}})^{2}$ $c^{2} (y - y_{0})^{2} - 2 a^{2} c (y - y_{0}) + a^{4} = a^{2} ((x - x_{0})^{2} + (y - y_{0} - c)^{2})$ $c^{2} (y - y_{0})^{2} - 2 a^{2} c (y - y_{0}) + a^{4} = a^{2} ((x - x_{0})^{2} + (y - y_{0})^{2} - 2 c (y - y_{0}) + c^{2})$ $c^{2} (y - y_{0})^{2} - 2 a^{2} c (y - y_{0}) + a^{4} = a^{2} (x - x_{0})^{2} + a^{2} (y - y_{0})^{2} - 2 a^{2} c (y - y_{0}) + a^{2} c^{2}$ $c^{2} (y - y_{0})^{2} - a^{2} (y - y_{0})^{2} - a^{2} (x - x_{0})^{2} = a^{2} c^{2} - a^{4}$ $(c^{2} - a^{2}) (y - y_{0})^{2} - a^{2} (x - x_{0})^{2} = a^{2} (c^{2} - a^{2})$

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

By definition we have $a^{2} = b^{2} + c^{2}$ , and thus $- b^{2} = c^{2} - a^{2}$ can be replaced and we obtain: $\frac{(y - y_{0})^{2}}{a^{2}} - \frac{(x - x_{0})^{2}}{- b^{2}} = 1 ⟹ \frac{(y - y_{0})^{2}}{a^{2}} + \frac{(x - x_{0})^{2}}{b^{2}} = 1$ .

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Example

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 $O Y$ . 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: $\frac{(y - y_{0})^{2}}{a^{2}} + \frac{(x - x_{0})^{2}}{b^{2}} = 1 ⟹ \frac{(y + 1)^{2}}{5^{2}} + \frac{(x - 1)^{2}}{4^{2}} = 1$