[Maths Class Notes] on Area of an Ellipse Pdf for Exam

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, 

(Image to be added soon)

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)

(Image to be added soon)

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