Area of Ellipse
The area of an ellipse is defined as the amount of region present inside it. Alternatively, the area of an ellipse is the total number of unit squares that can be fit in it. You might have observed many ellipseshaped shapes in our daily lives, for example, a cricket ground, a badminton racket, orbits of planets, etc. Let us explore a bit more about this shape while discussing its area, and the formula for the area of the ellipse with solved examples in the following sections.
1.  What Is the Area of Ellipse? 
2.  Area of Ellipse Formula 
3.  Proof of Formula of Area of Ellipse 
4.  Area of Ellipse Using Integration 
5.  How to Find the Area of Ellipse? 
6.  FAQs on Area of Ellipse 
What Is the Area of Ellipse?
The area or region covered by the ellipse in two dimensions is defined as the area of an ellipse. The area of an ellipse is expressed in square units like in^{2}, cm^{2}, m^{2}, yd^{2}, ft^{2}, etc. Ellipse is a 2D shape obtained by connecting all the points which are at a constant distance from the two fixed points on the plane. The fixed points are called foci. F_{1} and F_{2} are the two foci. As an ellipse is not a perfect circle, the distance from the center of the ellipse to the points on the circumference is not constant. So, an ellipse has two radii. The longest chord in the ellipse is called the major axis of the ellipse. The minor axis is the chord that is the perpendicular bisector to the major axis.
Ellipse is the locus of all the points, whose sum of distances from two fixed points on a plane is constant. For example, in the images given below the points P_{1}, P_{2 }are located in such a way that the sum of the distances of point P_{1} from the fixed points F_{1 }and F_{2} is equal to the sum of the distances of point P_{2} from the fixed points F_{1 }and F_{2}. That means if we join all such points P_{1}, P_{2, } P_{3} , etc; we will get a shape called an ellipse.
\(\begin{equation} \mathbf{P}_{1} \mathbf{F}_{1}+\mathbf{P}_{1} \mathbf{F}_{2}=\mathbf{P}_{2} \mathbf{F}_{1}+\mathbf{P}_{2} \mathbf{F}_{2}=\mathbf{P}_{3} \mathbf{F}_{1}+\mathbf{P}_{3} \mathbf{F}_{2}=...=\text{constant} \end{equation}\)
where,
 \(F_1, F_2\) = Two fixed points on the plane
 \(P_1, P_2, P_3\) ... = Points forming locus of ellipse
Area of Ellipse Formula
We can calculate the area of an ellipse using a general formula, given the lengths of the major and minor axis. The formula to find the area of an ellipse is given by,
Area of ellipse = π a b
where,
 a = length of semimajor axis
 b = length of semiminor axis
Proof of Formula of Area of Ellipse
Let E be an ellipse, with major axis of length 2a and minor axis of length 2b, aligned in a cartesian plane in the reduced form. Then, from the equation of ellipse in reduced form, we get,
\(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\)
Thus:
\(y = \pm \dfrac{b}{a} \sqrt{a^2  x^2}\)
Consider a circle, C, of radius 'a' whose center is at the origin. From the equation of circle formula with origin at center, we have,
\(x^2 + y^2 = a^2\)
⇒ y = ± \(\sqrt{a^2  x^2}\)
We observe from the formulas that each ordinate of the ellipse is \(b/a\) times the ordinate of the circle. The same thing is true for the vertical chords. We can thus also relate the areas of the two shapes as,
Area of ellipse = (b/a) × Area of circle
We know, area of a circle with the given equation, \(x^2 + y^2 = a^2\), is: A = πa^{2}, where 'a' is the radius of the circle.
⇒ Area of ellipse = (b/a)(πa^{2}) = πab
Area of an Ellipse Using Integration
The formula of area of an ellipse can also be derived using integration. The general equation for an ellipse is given as, \(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}} = 1\)
⇒ \(y = b.\sqrt{1\left ( \dfrac{x}{a} \right )^{2}}\)
We observe that the ellipse is divided into four quadrants. So, we can calculate the area of 1 quadrant and multiply by 4 to calculate the area of an ellipse.
Area, \(A = 4. \int_{0}^{a} y.dx\)
\(A = 4. \int_{0}^{a} \dfrac{b}{a} \sqrt{a^{2}x^{2}} dx\)
Substituting, x = a sin t, we get, dx = a cos t . dt
⇒ x = 0 changes to t = 0 and x = a changes to t = π/2
\(A = 4. \dfrac{b}{a}\int_{0}^{\dfrac{\pi}{2}} \sqrt{a^{2}a^{2} \sin^{2}t} .a \cos t .dt\)
⇒ \(A = 4b \int_{0}^{\pi/2}\sqrt{a^{2}(1sin^{2}t)} cost dt\)
⇒ \(A = 4ba \int_{0}^{\pi/2}\sqrt{cos^{2}t} cost dt\)
⇒ \(A = 4ba \int_{0}^{\pi/2}cos^{2}t dt\)
⇒ \(A = 4ba \int_{0}^{\pi/2}\dfrac{(1 + cos2t)}{2}dt\)
⇒ \(A = 4ab \left ( \dfrac{t}{2} + \dfrac{\sin 2t}{4} \right )_{0}^{\pi/2}\)
⇒ A = 4ab (π/4)
⇒ A = πab
where,
 a = length of semimajor axis
 b = length of semiminor axis
How to Find the Area of Ellipse?
The steps that can be followed to calculate the area of an ellipse using the length of the major and minor axis are given as,
 Step 1: Find the distance from the farthest point on the ellipse from the center ('a' i.e., length of the semimajor axis).
 Step 2: Find the distance from the closest point on the ellipse from the center ('b' i.e., length of the semiminor axis).
 Step 3: Multiply the product of a and b with π.
 Step 4: Mention the area in square units.
Important Notes
 Ellipse is the locus of all the points, whose sum of distances from two fixed points on a plane is constant.
 Area of ellipse = π a b
 Take the value of π as 3.14 or 22/7.
 Find the values of the semimajor axis (a) and the semiminor axis (b) and then use the formula for the area of an ellipse.
Solved Examples on Area of Ellipse

Example 1: The floor of a whispering gallery is constructed in the shape of an ellipse. Find its area, if the length of its major axis is 14 units and the length of its minor axis is 10 units.
Solution:
Given that,
Length of major axis = 14 units
⇒ length of semimajor axis is = (1/2) × 14 = 7 units.
Length of minor axis = 10 units
⇒ length of semiminor axis is = (1/2) × 10 = 5 units.Area of the ellipse is πab.
⇒ Area of whispering gallery = πab = π × 7 × 5 = (22/7) × 7 × 5 = 22 × 5 = 110 square units
Answer: Area of whispering gallery is 110 square units.

Example 2: Find the crosssectional area of football with its semimajor axis as 5.5 inches and semiminor axis as 3.5 inches using the area of an ellipse formula.
Solution:
American football is in the shape of an ellipse.
Length of semimajor axis = a = 5.5 inches
Length of semiminor axis = b = 3.5 inches
The area of the ellipse is,
πab = π × 5.5 × 3.5 = (22/7) × 5.5 × 3.5 = 11 × 5.5 = 60.5 square inches
Answer: The crosssectional area of football is 60.5 square inches.
FAQs on Area of Ellipse
What Is the Area of an Ellipse?
The area of an ellipse is the total region inside an ellipse in the twodimensional plane. It can be expressed in square units. Some common units used to express the area of an ellipse are in^{2}, cm^{2}, m^{2}, yd^{2}, ft^{2}, etc.
What Is the Formula to Find Area of an Ellipse?
The formula to calculate the area of an ellipse is given as, area of ellipse, A = πab, where, 'a' is the length of the semimajor axis and 'b' is the length of the semiminor axis.
How Do You Calculate Area of an Ellipse?
We can follow the steps given below to find the area of an ellipse,
 Step 1: Note the length of the semimajor axis, 'a', and length of the semiminor axis as 'b'.
 Step 3: Multiply the product of a and b with π.
 Step 4: Express the area in square units.
How to Derive the Formula for Area of an Ellipse?
Th formula of an ellipse can be derived using integration. We can calculate the area of any one quadrant of the ellipse with given major and minor axis and multiply the result by 4, to obtain the area of an ellipse. For detailed derivation, refer to section, Area of an Ellipse Using Integration.
What Is the Proof of Area of an Ellipse Formula?
The area of the ellipse formula can be proved using the analogy between the shape circle and ellipse. To see the detailed proof of formula using this method, refer to the section, Proof of Formula of Area of an Ellipse.
What Are the Units Used to Express the Area of an Ellipse?
In measurements, the area of an ellipse is expressed using square units or (unit)^{2}. The common units that can be used are square meters, square inches, square yards, etc.
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