Using our knowledge of conic sections from hereIntroduction To Conic Sections, in this chapter we will extend our understanding to cover the ellipse.

In our earlier section we saw that the when a conic section is formed using $0 \lt e \lt 1$ we get an ellipse. In general an ellipse can be viewed as an elongated circle, where the eccentricity $e$ is a measure of the elongation. When the elongation/eccentricity is $0$, the equation of an ellipse becomes a circle, while for $e = 1$, the elongation becomes infinite, that is the curve becomes open at the other end, giving us a parabola.

In other words, if you take any ellipse, and keeping one end fixed, if we are able to move the other end to infinity, that is the curve is unbounded on that side for all finite values, then we get a parabola with $e = 1$

Other than the concept of a double napped cone and a cutting plane, an ellipse can also be defined in two other ways using locus of a moving point on a plane.

- Ellipse is the locus of a point, the sum of whose distances from two fixed points are constant

- Ellipse is the locus of a point, whose distances from a fixed point and and a fixed straight line is in a constant ratio, less than $1$. This ratio is same as the eccentricity.

Derivation of the generic form of ellipse using the first definition.

We can derive the equation of the ellipse by choosing any two random points as focii, $F_1 = (x_1, y_1)$ and $F_2 = (x_2, y_2)$ on the Cartesian plane. This will give us the most generic form of equation for an ellipse.

But to arrive at the simplest form of equation, we can translate the axes so that the origin coincides with the  midpoint of the two focii, and then rotate the two axis such that the $x$-axis passes through the two focii.

In the new system, the two focii will become $F_1 = (-c, 0)$ and $F_2 = (c, 0)$. Let the distances of a point $P = (x, y)$ moving on the Cartesian plane from these two focii be $d_1$ and $d_2$, and $d_1 + d_2 = d$, where $d$ is constant for a given set of focii $F_1$ and $F_2$.

Before we proceed with the formal derivation of the equation, you can familiarize yourself with the ellipse using the widget below:

We know that the two distances, $d_1$ and $d_2$ are given by:

$d_1 = PF_1 = \sqrt{(x - (-c))^2 + (y - 0)^2} = \sqrt{(x  + c)^2 + (y - 0)^2}$, and

$d_2 = PF_2 = \sqrt{(x - c)^2 + (y - 0)^2}$

Using the condition of the locus we get: 

$\sqrt{(x  + c)^2 + y^2} + \sqrt{(x - c)^2 + y^2} = d$

$\Rightarrow \sqrt{(x  + c)^2 + y^2}  = d - \sqrt{(x - c)^2 + y^2}$

$\Rightarrow (x  + c)^2 + y^2 = d^2 + (x - c)^2 + y^2 - 2d\sqrt{(x - c)^2 + y^2}$

$\Rightarrow \cancel{x^2} + 2xc + \cancel{c^2} + \cancel{y^2} = d^2 + \cancel{x^2} - 2xc + \cancel{c^2} + \cancel{y^2}$

    $- 2d \sqrt{(x - c)^2 + y^2}$

$\Rightarrow 4xc = d^2 - 2d \sqrt{(x - c)^2 + y^2}$

$\Rightarrow 2d \sqrt{(x - c)^2 + y^2} = d^2 - 4xc$

$\Rightarrow 4d^2 \left[ (x - c)^2 + y^2 \right] = d^4 - 8d^2xc + 16x^2c^2$

$\Rightarrow 4d^2 (x^2 - 2xc + c^2 + y^2) = d^4 - 8d^2xc + 16x^2c^2$

$\Rightarrow 4d^2x^2 - \cancel{8d^2xc} + 4d^2c^2 + 4d^2y^2 = d^4 - \cancel{8d^2xc} + 16x^2c^2$

$\Rightarrow 4d^2x^2 + 4d^2c^2 + 4d^2y^2 = d^4 + 16x^2c^2$

$\Rightarrow 4d^2x^2 - 16x^2c^2 + 4d^2y^2 = d^4 - 4d^2c^2$

$\Rightarrow 4x^2(d^2 - 4c^2) + 4d^2y^2 = d^2(d^2 - 4c^2)$

$\Rightarrow \dfrac{4x^2\cancel{(d^2 - 4c^2)}}{d^2\cancel{(d^2 - 4c^2)}} + \dfrac{4\cancel{d^2}y^2}{\cancel{d^2}(d^2 - 4c^2)} = 1$

$\Rightarrow \dfrac{x^2}{\frac{d^2}{4}} + \dfrac{y^2}{\frac{(d^2 - 4c^2)}{4}} = 1$

Putting $y = 0$ in the above equation, we get:

$x = \pm \dfrac{d}{2}$, which means that the curve intersects the $x$-axis at $\pm \dfrac{d}{2}$.

If we substitute $\dfrac{d}{2} = a$, we get the length of the $x$ intercept as $a - (-a) = 2a$.

Similarly, putting $x = 0$, we get:

$y = \pm \dfrac{\sqrt{d^2 - 4c^2}}{2}$

We can substitute $\dfrac{\sqrt{d^2 - 4c^2}}{2} = b$, where $2b$ is the length of the $y$ intercept.

Therefore, we end up with the final standard form of ellipse as:

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

As an exercise, the reader should try to derive this form using the second definition mentioned above, in this section.

In this section we will formally introduce the definitions relevant to an ellipse. For all our definitions we will use the standard form of an ellipse which is $\dfrac{(x - x_0)^2}{a^2} + \dfrac{(y - y_0)^2}{b^2} = 1$.

Major & Minor Axes: The major and  minor axes are the largest and smallest line segments passing through the center and terminated by the ellipse on either sides. If in the standard form of equation, $a > b$, then, the major axis is the line passing through the center parallel to the $x$-axis, while the minor axis is the line passing through the center parallel to the $y$-axis. Else if, $a \lt b$, then the major axis is the line passing through the center and parallel to the $y$-axis and vice versa for the minor axis.

If $a > b$, then the equation of the major axis is $y = y_0$, for the standard form of ellipse above. Therefore, substituting $y = y_0$ in the equation of the ellipse, we get

$\dfrac{(x - x_0)^2}{a^2} + \dfrac{(y_0 - y_0)^2}{b^2} = 1$

$\Rightarrow \dfrac{(x - x_0)^2}{a^2} = 1$

$\Rightarrow x = x_0 \pm a$

Therefore, the intersection points of the ellipse with the major axis are $(x_0 + a, y_0)$ and $(x_0 - a, y_0)$

The length of the major axis is

$(x_0 + a) - (x_0 - a)$

$= x_0 + a - x_0 + a$

$= 2a$

Similarly, the length of the minor axis can be derived as $2b$ by substituting $x = x_0$. When $a < b$, the major axis will have a length of $2b$ and the minor axis will have a length of $2a$.

Vertex: The vertices of an ellipse are the points where the ellipse intersects the major axis. The coordinates of the vertices are $(x_0 \pm a, y_0)$ as we saw in the previous section.

Focus: The focii are a pair of points on the major axis at a distance of $ae$ on either side of the center. The coordinate of the focii are $(x_0 + ae, 0)$ and $(x_0 - ae, 0)$

Directrix: The directrices are a pair of lines parallel to the minor axis, lying outside the ellipse, at distances $\pm \dfrac{a}{e}$ from the center. Therefore, the equation of the two directrices will be $x = x_0 \pm \dfrac{a}{e}$

Eccentricity: The eccentricity, $(e)$, is defined as the ratio of the distance of any point on the ellipse from the focus and from the corresponding directrix, where $0 \lt e \lt 1$. For $e = 0$, the conic section becomes a circle, and for $e = 1$ the conic section is a parabola. Visually the eccentricity denotes the flatness of the ellipse. At zero eccentricity, there is no flatness and the two focii merge with the center. As the eccentricity increases the two focii move away from the center on either side. At $e = 1$ we get a parabola, which can also be visualized as an ellipse with the other end at infinity.

We can derive the expression for eccentricity using the following method. Let us consider an ellipse with the origin as its center, that is $x_0 = 0$ and $y_0 = 0$, then its focii are at $(\pm ae, 0)$ and its directrices are given by the equations $x = \pm \dfrac{a}{e}$.

Substituting $x = 0$ in the equation of the ellipse, we get $y = \pm b$.

Considering the point $(0, b)$ on the ellipse. Its distance from the focus $(ae, 0)$ is $\sqrt{a^2e^2 + b^2}$ and its distance from the directrix $x = \dfrac{a}{e}$ is $\dfrac{a}{e}$.

Therefore,

$\dfrac{\sqrt{a^2e^2 + b^2}}{\frac{a}{e}} = e$

$\Rightarrow \sqrt{a^2e^2 + b^2} = a$

$\Rightarrow a^2e^2 + b^2 = a^2$

$\Rightarrow a^2e^2  = a^2 - b^2$

$\Rightarrow e^2  = 1 - \dfrac{b^2}{a^2}$

$\Rightarrow e = \sqrt{1 - \dfrac{b^2}{a^2}}$

Latus Rectum: The latus rectum is a straight line perpendicular to the major axis, passing through the focus, and terminated on both sides by the elllipse.

We can use the following method to find the length of the latus rectum.

Since the length of the latus rectum is not dependent on the position of the center, we can assume that the ellipse is centered at the orgin, therefore having an equation of the form $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and has an eccentricity of $e$.

Therefore, the focii are at $(\pm ae, 0)$.

The equations of the latus rectum are given by $x = \pm ae$. We consider the right side latus rectum with the equation $x = ae$

To find the intersection point of this latus rectum we can substitute $x = ae$ in the equation of the ellipse, and get:

$\dfrac{a^2e^2}{a^2} + \dfrac{y^2}{b^2} = 1$

$\Rightarrow  \dfrac{y^2}{b^2} = 1 - e^2$

$\Rightarrow  \dfrac{y^2}{b^2} = 1 - e^2$

$\Rightarrow y = \pm b \sqrt{1 - e^2}$

Therefore, the two intersection points of the latus rectum with the ellipse are $(ae, b \sqrt{1 - e^2})$ and $(ae, -b \sqrt{1 - e^2})$.

Therefore, the length of the latus rectum is $2b \sqrt{1 - e^2}$

Substituting $e = \sqrt{1 - \dfrac{b^2}{a^2}}$ we get the length as:

$l = 2b \sqrt{1 - \left(1 - \dfrac{b^2}{a^2}\right)}$

$\Rightarrow l = 2b \sqrt{\dfrac{b^2}{a^2}}$

$\Rightarrow l = \dfrac{2b^2}{a}$

A ray of light passing through any one of the focii, after being reflected from the inner surface of an ellipse will pass through the other focii, as shown in the following property.

This, in other words, means that the normal to an ellipse at any point bisects the angle formed by the lines joining that point and the two focii.

The area of an ellipse with $2a$ and $2b$ as the major and minor axis is given by $\pi a b$. This relation can be derived using integration.

If we have an ellipse centered at the origin, represented by $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and replace $x = a\cos t$ and $y = b \sin t$, then for any value of $t$, we get the $LHS$ as:

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}$

$= \dfrac{a^2 \cos^2 t}{a^2} + \dfrac{b^2 \sin^2 t}{b^2}$ 

$= \cos^2 t + \sin^2 t$

$= 1 = RHS$

Therefore,

$x = a\cos t$

$y = b \sin t$

satisfy the equation of the ellipse for all values of $t$, and gives us the parametric form of the ellipse.