How To Find The Area Of An Ellipse

Ellipse is a 2-dimensional shape. Information technology is an integral office of the conic section. It is a curve on a airplane in which the sum of the altitude to its two focal points or foci is always a constant quantity from the specified point. Ellipse is from a family of circles with two focal points. The eccentricity value of the ellipse is ever less than 1 and the general equation of the ellipse is given by

(x²/aⁱⁱ) + (y²/b^two) = i

Where a represents the length of the semi-major axis and b represents the length of the semi minor-axis.

Parts of Ellipse

Some of the important parts of the ellipse are:

Focus: Ellipse has 2 foci or focal points whose coordinates are F1(q, 0), and F2(-q, 0) And the distance between them is 2q
Heart: It is the midpoint of the line joining two foci.
Major axis: It is the longest diameter of the ellipse. Or we can say that a line segment that connects the two farthest points present on the ellipse passes through the centre of the ellipse.
Minor centrality: It is the shortest diameter of the ellipse. Or nosotros tin can say that it is a line segment that joins the two nearest points present on the ellipse and passes through the centre of the ellipse.
Latus Rectum: Information technology is a line that is fatigued perpendicular to the transverse centrality and passes through the foci.

Area of Ellipse

An ellipse is created by connecting all the points that are present at a constant distance from two fixed points in the plane. Here the two stock-still points are known as foci. So the area of the ellipse is known as the total corporeality of area present in information technology. It is represented past cmⁱⁱ, in^two, m^two, etc. The area of the ellipse can be calculated using the length of the semi-major axis and the length of the semi-minor axis. The area of the ellipse is given by-product of the length of the semi-major axis and minor centrality with π (22/7). The formula is given by

Expanse of ellipse A = π × chiliad × n

Where chiliad is the length of the semi-major axis and n is the length of the semi-minor axis.

Proof of Area of Ellipse formula

Let us consider A be an ellipse, 2m is the major axis and 2n be the small axis. It is aligned in the cartesian plane in the reduced form and then the equation of the ellipse is

xⁱⁱ/m² + y²/n² = 1

So,

y = ±due north/1000√kⁱⁱ – x²……(1)

Now assume a circle of radius m and whose middle lies at the origin. So the equation of the circumvolve is

xⁱⁱ + yⁱⁱ = thouⁱⁱ

And so, y = ±√one thousand² – x² ……(2)

On comparing equations (1) and (2) we conclude that the ellipse is n/1000 times the circle.

So nosotros tin write the surface area of the two shapes equally:

Area of ellipse = n/chiliad x Area of the circumvolve

So,

Area of ellipse = n/1000 x (πmⁱⁱ) = πnm

Hence proved

Steps to find the area of Ellipse

Step 1: Find the length of the semi-major axis or the distance from the farthest signal from the center that is m.

Step two: Find the length of the semi-modest axis or the distance from the nearest point from the center that is n.

Stride 3: Now put all these values in the area formula to calculate the area.

Pace 4: At present apply the units of surface area.

Case: A floor of a room is constructed in the shape of an ellipse whose major axis length is xv cm and modest axis length is eleven cm. Now find the surface area of the ellipse.

Solution:

Given that

Major axis length(thousand) = 15 cm

minor axis length(northward) = 11 cm

Area of ellipse (A) = π x m x n

= 22/7 ten xv x xi

= 22/7 x 165

= 518.57 cm²

Finding the expanse of an ellipse using integration

Nosotros can also find the expanse of an ellipse using integration. As we know the full general equation of the ellipse is

x²/mⁱⁱ + y²/n² = 1

y = ±n/m√m² – x²……(1)

As we know that the ellipse is divided into four quadrants so we find the area or one quadrant so multiple with 4 to get the surface area of the full ellipse

And then, A = 4 $\int_{m}^{0} y \,dx$

= 4 $\int_{0}^{m} \frac{n}{m}\sqrt{m^2-x^2} \,dx$

= 4n/thousand $\int_{0}^{m} \sqrt{m^2-x^2} \,dx$

Now put x = yard sint, dx = yard cost.dt and t = 0 and t = π/2 in the in a higher place equation

A = 4n/grand $\int_{0}^{\pi/2} \sqrt{m^2-m^2sin^2t} \,.mcost.dt$

= 4n $\int_{0}^{\pi/2} \sqrt{m^2(1-sin^2t)} \,costdt$

= 4mn $\int_{0}^{\pi/2} \sqrt{cos^2t} \,costdt$

= 4mn $\int_{0}^{\pi/2} cos^2t \,dt$

= 4mn $(\frac{t}{2} + \frac{sin2x}{4})^{\pi/2}_0$

= 4mn(π/4)

= πmn

Sample Questions

Question 1: What is the surface area of an ellipse if the length of the semi-major axis and the semi-small centrality is 6cm, 3cm respectively.

Solution:

Given

Length of semi major centrality (a) = 6cm

Length of semi minor axis (b) = 3cm

Surface area of ellipse= π x a ten b

= (22/vii) x 6 x iii

= 56.57 cm²

So area of given ellipse is 56.57 cmⁱⁱ.

Question 2: What is the area of ellipse if the length of the semi-major axis and semi-minor axis is 10cm, four.5cm respectively.

Solution:

Given

Length of semi major axis (a) = 10cm

Length of semi small-scale axis (b) = iv.5cm

Expanse of ellipse= π x a x b

= (22/7) 10 10 x 4.5

= 141.42 cmⁱⁱ

And then expanse of given ellipse is 141.42 cmⁱⁱ.

Question 3: What is the area of the ellipse if the length of major axis and minor axis is 10cm, 5cm respectively.

Solution:

Given

Length of major axis = 10cm

Length of minor centrality = 5cm

So we demand to discover semi major axis and semi pocket-size axis length.

Length of semi major axis (a) = Length of major axis/2

= 10/2 = 5cm

Length of semi small-scale axis (b) = Length of minor axis/ii

= 5/two = two.5cm

Area of ellipse = π x a 10 b

= (22/7) x 5 10 2.five

= 39.25 cm^two

So expanse of given ellipse is 39.25 cm^two.

Question 4: Detect the surface area of ellipse if the length of major axis and small axis is 12cm, 4cm respectively.

Solution:

Given

Length of major axis = 12cm

Length of small-scale centrality = 4cm

So nosotros need to find semi major axis and semi minor axis length.

Length of semi major axis (a) = Length of major axis/2

= 12/ii = 6cm

Length of semi minor axis (b) = Length of minor axis/2

= 4/2 = 2cm

Area of ellipse = π ten a x b

= (22/7) x half dozen x 2

= 37.71 cm²

So area of given ellipse is 37.71 cm².

Question five: Observe the expanse of ellipse if the length of major axis and semi minor axis is 8cm, 2.5cm respectively.

Solution:

Given

Length of major axis = 8cm

Length of semi minor axis (b) = 2.5cm

And then we demand to find length of semi major axis

Length of semi major axis (a) = Length of major axis/2

= 8/2 = 4cm

Area of ellipse = π ten a 10 b

= (22/7) 10 4 x ii.5

= 31.43 cm²

So expanse of given ellipse is 31.43 cmⁱⁱ.