Like parabola, hyperbola and circle, an ellipse is also a conic section. Ellipse is a closed two-dimensional curve which is obtained when a cone is intersected by a plane that is not parallel to its base, as shown in the following figure:

Circle is a special case of ellipse. When the intersecting plane is parallel to the base of the cone, a circle is formed. We can think of an ellipse as a circle which is pressed at the top. An ellipse is a locus of a point, sum of whose distances from two fixed points inside the elliptic curve is a constant.

Astronomers, initially, had an impression that the orbits of planets are oval, but later Kepler discovered that orbits of each planet is an ellipse and sun is at one of its foci.

## Standard Form of an Ellipse

Standard forms of horizontal ellipses can be expressed as two equations below:
• When center of the ellipse is located at origin (0, 0):

• When center of the ellipse is located at a point (h, k):

Standard forms of vertical ellipses can be expressed as two equations below:
• When center of the ellipse is located at origin (0, 0):
When center of the ellipse is located at a point (h, k):
Where,
a = Length of semi-major axis,
b = Length of semi-minor axis,
(Provided that a > b),
h and k are coordinates of center (h, k),
x and y are coordinates of an arbitrary point (x, y).

## Properties of an Ellipse

Following two images are demonstrating horizontal and vertical ellipses:
• Horizontal Ellipse
• Vertical Ellipse

Important properties of a ellipse with center at origin are listed below:
• The intersecting point of ellipse at X-axis is called vertex. An ellipse has two vertices. The coordinates of vertices are (a, 0) and (- a, 0).
• The intersecting point of ellipse at Y-axis is called co-vertex. An ellipse has two co-vertices. The coordinates of co-vertices are (0, b) and (0, - b).
• There are two foci whose coordinates are (ae, 0) and (- ae, 0).
• Length of major axis is "2a" and that of minor axis is "2b".
• Eccentricity of an ellipse is the ratio of distance of center to focus and length of semi-major axis and is denoted by "e". For ellipse, e < 1.
• An ellipse has two directrices, whose equations are $x=$$\frac{a}{e} and x=-$$\frac{a}{e}$
• The line segment inside the ellipse passing through focus is called latus retum whose length is $\frac{2b^{2}}{a}$.
• Area enclosed by an ellipse = $\pi$ab.

## Formula for an Ellipse

Equation of an ellipse centered at origin is given below:
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$=1 Equation of an ellipse centered at a point (h, k) is given below: \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}$$=1$

Eccentricity of an ellipse can be deduced from the following formulas:
1. $e=$$\sqrt{1-\frac{b^{2}}{a^{2}}} 2. e=$$\frac{c}{a}$
Where,
a = Length of semi-major axis
b = Length of semi-minor axis
c = Length of focus from center.
For ellipse, e < 1.
Formula for tangent at a point (x1, y1) at an ellipse is given below:
$\frac{xx_{1}}{a^{2}}+\frac{yy_{1}}{b^{2}}$$=1 ## Finding the Equation of an Ellipse Sometimes, equation of an ellipse is not given in standard form. In this case, firstly, we need to deduce the equation in standard from. Few examples of finding the equation of an ellipse are given below: ### Solved Examples Question 1: Deduce the equation of ellipse from the equation 5x^{2}+3y^{2}=15. Solution: 5x^{2}+3y^{2}=15 Divide the equation by 15 on both the sides to get 1 in right hand side of equation: \frac{5x^{2}}{15}+\frac{3y^{2}}{15}$$=1$

$\frac{x^{2}}{3}+\frac{y^{2}}{5}$$=1 \frac{x^{2}}{(\sqrt{3})^2}+\frac{y^{2}}{(\sqrt{5})^2}$$=1$

Which is standard equation of an ellipse centered at origin with a = √3 and b = √5.

Question 2: Find the standard form of an ellipse from equation 2x2 + y2 - 4x + 8y + 6 = 0.
Solution:

Using complete-the-perfect-square method,
2 (x2 - 2x) + (y2 + 8y) + 6 = 0
2 (x2 - 2x + 1) - 2 + (y2 + 8y + 16) - 16 + 6 = 0
2 (x - 1)2 + (y + 4)2 - 12 = 0
2 (x - 1)2 + (y + 4)2 = 12

$\frac{(x-1)^{2}}{6}+\frac{(y+4)^{2}}{12}$$=1 \frac{(x-1)^{2}}{(\sqrt{6})^2}+\frac{(y-(-4))^{2}}{(2\sqrt{3})^2}$$=1$

Which is standard equation of an ellipse centered at (1, - 4) with a = √6 and b = 2√3.

## Graphing an Ellipse

Following steps should be followed while graphing an ellipse:
1. Deduce the given equation into standard form of an ellipse (if not).
2. Determine whether it is a horizontal ellipse or a vertical ellipse.
3. Find the coordinates of center.
4. Find the lengths of major and minor axes.
5. Determine the vertices and co-vertices.
6. Calculate the eccentricity e.
7. Find the coordinates of foci.
8. By using the above information, plot the corresponding points and plot a graph of ellipse of given equation.
Let us consider one example:

### Solved Example

Question: Sketch a graph for the following equation of ellipse:
9x2 + 25y2 - 225 = 0.
Solution:

Deduce the given equation in standard form of ellipse:
9x2 + 25y2 - 225 = 0
9x2 + 25y2 = 225

$\frac{x^{2}}{25}+\frac{y^{2}}{9}$$=1 \frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}$$=1$

Since bigger value is below x, hence this is a horizontal ellipse.
Center is at (0, 0).
a = 5 and b = 3
Vertices are (5, 0) and (- 5, 0).
Co-vertices are (0, 3) and (0, - 3).

Eccentricity $e=$$\sqrt{1-\frac{b^{2}}{a^{2}}} e=$$\sqrt{1-\frac{9}{25}}$$=$$\frac{4}{5}$

e = 0.8
Coordinates of foci are (4, 0) and (- 4, 0).
Using the information obtained, we get the following ellipse:

## Ellipse Equation Examples

Few examples based on ellipse equations are as follows:

### Solved Examples

Question 1: Consider an ellipse 4x2 + 3y2 =12. What kind of ellipse is this? Determine coordinates of vertices and foci. Find the area enclosed by the ellipse.
Solution:

4x2 + 3y2 =12
Dividing both the side by 12, we get:

$\frac{x^{2}}{3}+\frac{y^{2}}{4}$$=1 \frac{x^{2}}{(\sqrt{3})^{2}}+\frac{y^{2}}{2^{2}}$$=1$

Since bigger value is below y, hence this is a vertical ellipse.
a = 2 and b = √3
Coordinates of vertices: (0, 2) and (0, - 2)

$e=$$\sqrt{1-\frac{b^{2}}{a^{2}}} e=$$\sqrt{1-\frac{3}{4}}$

$e=$$\frac{1}{4} = 0.5 Coordinates of foci: (0, 2) and (0, - 2) Area = π a b = π \times 2 \times √3 = 10.8828 (approx) Question 2: Deduce the following equation of ellipse in standard form and find the center and eccentricity: x2 + 2y2 - 2x - 2y + 1 = 0 Solution: Given equation: x2 + 2y2 - 2x - 2y + 1 = 0 (x2 - 2x) + 2(y2 - y) + 1 = 0 (x2 - 2x + 1) + 2(y2 - y + \frac{1}{4}) - \frac{1}{2}= 0 (x - 1)2 + 2(y - \frac{1}{2})2 = \frac{1}{2} \frac{(x-1)^{2}}{(\frac{1}{\sqrt{2}})^{2}}+\frac{(y-\frac{1}{2})^{2}}{(\frac{1}{2})^{2}}$$=1$

Center = (1, $\frac{1}{2}$)

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