Equation of the ellipse with foci on the axis OX

An ellipse is the geometric place of the points in the plane which sum of distances to two fixed points, called foci, is constant.

We will suppose that in this case the foci $F$ and $F^{'}$ are on the axis $O X$ , so that they are defined by $F^{'} = (- c, 0)$ and $F = (c, 0)$ and therefore the ellipse is centred on the origin.

This way, for the ellipse definition we will write that for any point $P$ of the ellipse, the following is satisfied: $\overset{―}{P F} + \overset{―}{P F^{'}} = 2 a$ 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:

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Let's develop it: $\overset{―}{P F} + \overset{―}{P F^{'}} = 2 a$ which is equivalent to the expression: $\sqrt{(x - c)^{2} + y^{2}} + \sqrt{(x + c)^{2} + y^{2}} = 2 a$ So, first, we move the second root to the other side of the equality: $\sqrt{(x - c)^{2} + y^{2}} = 2 a - \sqrt{(x + c)^{2} + y^{2}}$

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

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

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

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

Now we divide both sides of the expression by the factor $a^{2} b^{2}$ and it gives: $\frac{b^{2} x^{2} + a^{2} y^{2}}{a^{2} b^{2}} = \frac{a^{2} b^{2}}{a^{2} b^{2}}$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ The latter expression is the equation of the ellipse that we wanted to find.

Example

If they give us the expression $\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1$ then this corresponds to an ellipse centred on the origin for which we calculate the semiaxes in the following way: $a^{2} = 25 \Rightarrow a = \sqrt{25} = 5$ $b^{2} = 4 \Rightarrow b = \sqrt{4} = 2$

What is the value of the semiaxes of the ellipse $\frac{x^{2}}{3} + \frac{y^{2}}{8} = 1$ ?

Equaling the denominators to the squares of the above mentioned lengths we obtain: $a = \sqrt{3} b = \sqrt{8} = 2 \sqrt{2}$

Now we are going to work a little with this equation.

Example

We are going to find the typical elements and the limited equation of the ellipse of foci: $F^{'} (- 3, 0)$ and $F (3, 0)$ , such that its biggest axis measures $10$ .

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Since the biggest axis measures $10$ we know that the biggest semiaxis will be half.

This way, we obtain: $2 a = 10 \Rightarrow a = 5$ .

Since we know that the foci are the points $F^{'} (- 3, 0)$ and $F (3, 0)$ , the distance between them is $6$ .

Therefore: $2 c = 6 \Rightarrow c = 3$ .

Since we know the relation $a^{2} = b^{2} + c^{2}$ , by isolating $b$ of the above mentioned equation we obtain: $b^{2} = 5^{2} - 3^{2} = 25 - 9 = 16 \Rightarrow b = 4$

Now , since we already know the major and minor semiaxes, we take the equation of the ellipse and replace the values in it, obtaining in this way the equation of this ellipse. $\frac{x^{2}}{5^{2}} + \frac{y^{2}}{4^{2}} = 1$ Finally, we can calculate the eccentricity, which is $e = \frac{c}{a} = \frac{3}{5}$