Area of an Ellipse

An ellipse is a two-dimensional shape that you must have encountered in your geometry class. This shape looks like a flat, elongated circle. Calculating the area of an ellipse is easy when you know the measurements of the major radius and minor radius.

It is basically a plane curse with two focal points, such that the sum of the distance from these focal points to anywhere on the circumference is always constant.

For example: - Consider an Ellipse with two focal point F₁ and F₂ 

Now consider two different points A and B on the circumference of the ellipse then, 

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F₁ A +F₂ A = ∝

F₁ B + F₂ B = ∝

Where α is the constant. F₁ and F₂ are the focus of Ellipse.

For Horizontal Major Axis-

x2 /a2 + y2 /b2 = 1, (where a>b)

Or, y=b.1−(xa)²−−−−−−−√

For Vertical Major axis-

x2 /b2 +y2 /a2 = 1, (where a>b)

Or, y=a.1−(xb)²−−−−−−−√

Some Other Terminology

Consider an Ellipse with 

Point F₁ and F₂ are the two focal points of the Ellipse, the line joining the two focal points and cutting on the circumferences is called the Vertex. (R1 and P)

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O is the midpoint of PR and is the center of the Ellipse. A line when drawn perpendicular to this center point O gives the minor axis of the Ellipse.

Ellipse & its Formulas

Formula of the are of the ellipse

The area of an Ellipse can be calculated by using the following formula

Area =  π * r₁ * r₂

Where r₁ is the semi-major axis or longest radius and r₂ is the semi-minor axis or smallest radius.

The area is all the space that lies inside the circumference of the Ellipse.

Steps Involved in Calculating the Area

• Find the major radius of the Ellipse

• Calculate the minor radius 

• Multiply both the radius with to calculate the area

• Area = π * r₁ * r₂

Derivation of Equations of Ellipse

Now, we take a point P (x, y) on the ellipse such that, PF1 + PF2 = 2a

By the distance formula, we have,
√ {(x + c)2 + y2} + √ {(x – c)2 + y2} = 2a
Or, √ {(x + c)2 + y2} = 2a – √ {(x – c)2 + y2}

Further, let’s square both sides. Hence, we have
(x + c)2 + y2 = 4a2 – 4a√ {(x – c)2 + y2} + (x – c)2 + y2

Simplifying the equation, we get √ {(x – c)2 + y2} = a – x(c/a)
Now, by squaring both the sides and simplifying it we get, x2/a2 + y2/ (a2 – c2) = 1

We know that c2 = a2 – b2. Therefore, we have x2/a2 + y2/b2 = 1
Therefore, we can say that any point on the ellipse satisfies the equation:

x2/a2 + y2/b2 = 1 … (1)

Converse Situation

Let’s look at the converse situation now. If P (x, y) satisfies equation (1) with 0 < c < a, then y2 = b2(1 –(x2/a2))

Therefore, PF1 = √ {(x + c)2 + y2}
= √ {(x + c)2 + b2(1-(x2/a2))}

Let us now simply this equation and also substitute b2 with a2 – c2. By this, we get PF1 = a + x(c/a)
Using similar calculations for PF2, we get PF2 = a – x(c/a)
Hence, PF₁ + PF₂ = {a + x(c/a)} + {a – x(c/a)} = 2a.

Therefore, any point that satisfies equation (1), i.e. x2/a2 + y2/b2 = 1, signifies that it is positioned on the ellipse. Also, the equation of an ellipse with the centre of the origin and major axis along the x-axis is:

x2/a2 + y2/b2 = 1.

Note: Solving the equation (1), we get

x2/a2 = 1 – y2/b2 ≤ 1

Therefore, x2 ≤ a2. So, – a ≤ x ≤ a. Hence, it can be concluded that the ellipse is lying between lines x = – a and x = a and touches these lines. Its equation {Fig. 5 (b)} is:

x2/b2 + y2/a2 = 1.

Hence the Standard Equations of Ellipses are:

x2/a2 + y2/b2 = 1.

x2/b2 + y2/a2 = 1.

Observations

An ellipse shows symmetry with respect to both coordinate axes. In simple words, if (m, n) is a point on the ellipse, then (- m, n), (m, – n) and (- m, – n) also fall on it.

The foci always lie on the major axis.

If the coefficient of x2 has a larger denominator, then the major axis is along the x-axis.

If the coefficient of y2 has a larger denominator, then the major axis is along the y-axis.

What are the applications of Ellipse in real life?

The ellipse has a close reference with football when it is rotated on its major axis. 

Another classic example is the orbit of planet Pluto. 

Solved Examples 

Q 1: Find out the coordinates of the foci, vertices, lengths of major and minor axes, and the eccentricity of the ellipse 9x2 + 4y2 = 36.

A: Given, 9x2 + 4y2 = 36. The first step is to divide both the LHS and RHS by 36, which gives us:

x2/4 + y2/9 = 1

We can notice that the denominator of y2 is larger than that of x2. Hence, the major axis is along the y-axis. The next step is to compare it with the standard equation. By comparing them we have, a2 = 4 or a = 2 and b2 = 9 or b = 3

Also, c2 = a2 – b2
Or, c = √ (a2 – b2) = √ (9 – 4) = √5
And, e = c/a = √5/3

Henceforth,

The foci are (0, √5) and (0, – √5).

Vertices are (0, 3) and (0, – 3)

Length of the major axis = 6

Length of the minor axis = 4