Content Curator
Area of Ellipse is a x b x π. In this calculation, two units of length are multiplied together, which results in output of units squared. For example, an ellipse has a major radius: 5 units and a minor radius: 3 units, area of ellipse would be 3 x 5 x π, or about 47 square units.
| Table of Content |
Also read: Definite Integral Formula
What is Ellipse?
[Click Here for Sample Questions]
Ellipse is the set of all points in a plane, the sum of whose distances from two distinct fixed points (foci) is constant. The line through the foci intersects the ellipse at two points called vertices. The chord joining the vertices is the major axis, and its midpoint is the centre of the ellipse.
The chord perpendicular to the major axis at the centre is the minor axis of the ellipse. The centre of an ellipse is the midpoint of both the major and minor axis. The axis are perpendicular at the centre. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci.
The length of the major axis is denoted by 2a, the length of the minor axis by 2b and the distance between the foci by 2c. Thus, the length of the semi major axis is a and semi-minor axis is b
Standard Equation of ellipse: \(\frac{x^2}{a^2}+ \frac{y^2}{b^2}\) = 1
From the standard equations of the ellipse we observe that:
- Ellipse is symmetric with respect to both the coordinate axes since if (x, y) is a point on the ellipse, then (– x, y), (x, –y) and (– x, –y) are also points on the ellipse.
- The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of x2 has the larger denominator and along the y-axis if the coefficient of y2 has the larger denominator.
The area of ellipse is given by:
(Area= π ab), Where: a (horizontal segment) = Length of semi-major axis
b (vertical segment) = length of semi-minor axis
Discover about the Chapter video:
Conic Sections Detailed Video Explanation:
Steps to Find the Area of Ellipse
[Click Here for Sample Questions]
The steps to calculate the area of an ellipse by using the length of the major and minor axis are:
Step 1: Calculate ('a' i.e., length of the semi-major axis) the distance from the farthest point on the ellipse from the center.
Step 2: Calculate ('b' i.e., length of the semi-minor axis) the distance from the nearest point on the ellipse from the center.
Step 3: Multiply π with the product of a and b.
Step 4: We get the required area in square units.
Also read: Area of a Triangle
Proof of Area of Ellipse
[Click Here for Sample Questions]
The general equation for an ellipse is
\(\frac{x^2}{a^2}+ \frac{y^2}{b^2}\) = 1 , y = \(b.\sqrt{1-(\frac{x}{a})^2}\), y = \(\frac{b}{a} \sqrt{a^2–x^2}\)
The standard equations of ellipses have centre at the origin and the major and minor axis are coordinate axes.
For Horizontal Major Axis:
x2 /a2 + y2 /b2 = 1, (where a>b)
For Vertical Major Axis
x2 /b2 +y2 /a2 = 1, (where a>b)
The ellipse is divided into four quadrants. So we get the area of an ellipse by calculating the area of one quadrant, multiplied by 4.
Area of one quadrant=
A= \(4. \int^a_0 y.dx\)
A= \(4.\int_0^a b/a \sqrt{a^2-x^2 }dx\)
A = \(4.\frac{b}{a} \int_0^a.1. \sqrt{a^2-x^2} dx\)
Now by substituting x = a sin t
dx = a cos t . dt (x = 0 changes to t = 0 and x = a changes to t = π/2)
Area of ellipse=πab, where, a = length of semi-major axis and b = length of semi-minor axis
By analysing the formula, we can see that each ordinate of the ellipse is b/a times the ordinate of the circle and the vertical chords. Thus we can relate the areas of ellipse and circle as,
Area of ellipse = (b/a) × Area of circle
We know that, area of a circle with the equation, x2 + y2 = a2 , is A = πa2 where 'a' is the radius of the circle.
⇒ So, Area of ellipse = (b/a)(πa2) = πab
Special Case of Ellipse
A circle is a unique case of an ellipse. The formula used for the area of a circle is π r2. However, representing that a circle is an ellipse with equal minor and major axes(a=b) , the formula for the area of the ellipse becomes the same as the formula for area of a circle.
Also read: Trapezoid Formula
Eccentricity
[Click Here for Sample Questions]
One of the reasons why it was difficult for early astronomers to detect that planetary orbits are ellipses is that the foci of the planetary orbits are relatively close to their centers, and therefore the orbits are almost circular.
Definition of Eccentricity: The eccentricity (e) of an ellipse is given by the ratio e= c/a, (where c= distance of the focus from the center of the ellipse and a= length of major axis)
Note: 0 < e < 1 every ellipse.
To see how this ratio is used to describe the shape of an ellipse, note that because the foci of an ellipse are located along the major axis between the vertices and the center, it follows that: 0<c<a
For an ellipse that is nearly circular, the foci are close to the center and the ratio c/a is small. On the other hand, for an elongated ellipse, the foci are close to the vertices, and the ratio c/a is close to.
Also read: Quadrilateral Formula
Latus Rectum
[Click Here for Sample Questions]
Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse.
Length of the latus rectum of the ellipse(\(\frac{x^2}{a^2}+ \frac{y^2}{b^2}\)=1)
Let the length of AF2 be l.
The coordinates of A are (c, l ),i.e., (ae, l ) Since A lies on the ellipse \(\frac{x^2}{a^2}+ \frac{y^2}{b^2}\)=1)
\(\frac{ae^2}{a^2}+ \frac{y^2}{b^2}\)=1
l2 = b2 (1-e2)
Now, we know that e = c/a
e2= c2/a2 = (a2-b2)/ a2 = 1- b2/a2
Therefore, l2=b4/a2, i.e. l=b2/a
As, the ellipse is symmetric with respect to y-axis (of course, it is symmetric w.r.t. both the coordinate axes), AF2 = F2B and so length of the latus rectum is 2b2/ a.
Applications
[Click Here for Sample Questions]
Ellipses have many practical and aesthetic uses like:
- Machine gears, supporting arches, and acoustic designs often involve elliptical shapes.
- The elliptical orbit of the moon about Earth and the orbits of satellites and planets are also ellipses.
- To find the area of oval fields like football field.
- To find the area of oval structures like pool, table, floor etc.
Also read: Calculus Formula
Things to Remember
- Ellipse is the locus of all the points.
- Area of ellipse = a b, (π = 3.14 or 22/7) .
- a= semi-major axis and b= the semi-minor axis.
- a is the distance from the farthest point on the ellipse from the center.
- b the distance from the nearest point on the ellipse from the center.
- Area of ellipse is b/a times the area of circle.
Also read: Determinant Formula
Sample Questions
Ques. Find the equation of ellipse given that foci and eccentricity are (±2,0) and 1/ 2 respectively. (2 marks)
Solution: Let the equation of the ellipse be \(\frac{x^2}{a^2}+ \frac{y^2}{b^2}\)=1. Then, coordinates of the foci are (±ae,0).
Therefore, ae = 2 ⇒ a = 4
We have b2 = a2(1–e2) ⇒ b2 =12
Thus, the equation of the ellipse is x2/16 + y2/12 = 1
Ques. Determine major axis and its length for the following ellipse? (1/9) x 2 + (9/25) y 2 = 1/25 (2 marks)
Solution: Multiply the equation by 25:
(25/9) x 2 + (9) y 2 = 1
Write above equation in the form x 2 / a 2 + y 2 / b 2 = 1:
x 2 / (3/5) 2 + y 2 / (1/3) 2 = 1
with a = 3/5 and b = 1/3.
The major axis is the x axis and length is equal to 2a = 6/5 = 1.2
Ques. If an ellipse is given by the following equation: 8x 2 + 2y 2 = 32 Find a) the major axis and the minor axis and their lengths, b) the vertices of the ellipse, c) and the foci of the ellipse. (2 marks)
Solution: a) Divide all terms of the equation by 32 to obtain
x 2 / 4 + y 2 / 16 = 1
Write the above equation as:
x 2 / b 2 + y 2 / a 2 = 1
with a = 4 and b = 2 and a > b.
Since the major axis is the y axis and the minor axis is the x axis.
The length of the major axis = 2a = 8
The length of the minor axis = 2b = 4
b) The vertices are on the major axis at the points (0 , a) = (0 , 4) and (0 , -a) = (0 , -4)
c) The foci are on the major axis at the points (0 , c) and (0 , -c) such that
c 2 = a 2 - b 2 = 12.
Hence the foci are at the points (0 , 2√3) and (0 , -2√3)
Ques. Find the equation of the ellipse whose center is the origin of the axis and has a focus at (0 , -4) and a vertex at (0 , -6). (2 marks)
Solution: Both the focus and the vertex lie in the y axis which means that the major axis is the y axis. The equation of the ellipse has the form
x 2 / b 2 + y 2 / a 2 = 1
a = the distance from the center of the ellipse to the a vertex and is equal to 6.
c = the distance from the center of the ellipse to a focus and is equal to 4.
a, b and c are related as:
b 2 = a 2 - c 2 = 36 - 16 = 20
b = 2√5
Thus, the equation of the ellipse is given by:
x 2 / 20 + y 2 / 36 = 1
Ques. If 3x + 4y = 12√2 is a tangent to the ellipse x2/a2 + y2/9 = 1, for some a ? R then the distance between the foci of the ellipse is: (4 marks)
Solution: Here 3x + 4y = 12√2 is the tangent to ellipse.
=> y = -3x/4 + 3√2 ..(1)
We know, the equation of tangent to ellipse x2/a2 + y2/b2 = 1 is y = mx + √(a2m2 + b2)
Here b2 = 9
So, equation of tangent to given ellipse is y = mx + √(a2m2 + 9)
Now, using (1), we have
m = -3/4 and √(a2m2 + 9) = 3√2
=> (a2m2 + 9) = 18
=> (a2(-3/4)2 + 9) = 18
=> a2 x 9/16 = 9
=> a = 4
Now, e2 = 1 – b2/a2 = 1 – 9/16 = 7/16
Or e = √7/4
Distance between foci: 2ae = 2 x 4 x √7/4 = 2√7
Ques. The ellipse x2 + 4y2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through the point (4, 0). Find the equation of the ellipse. (2 marks)
Solution: Given equation of ellipse : x2 + 4y2 = 4
x2/4 + y2/1 = 1
Here a = 2 and b = 1
We know the general equation of ellipse is x2/ a2+ y2/b2 = 1
4/16 + 1/b2
=> b2 = 4/3
Therefore, x2/16 + y2/(4/3) = 1
=> x2/16 + 3y2/4 = 1
=> x2 + 12 y2 = 16
Ques. An ellipse is given by the equation (x - 1) 2 / 9 + (y + 2) 2 / 16 = 1 Find a) centre, b) major and minor axes and their lengths, c) vertices, d) foci. (2 marks)
Solution: Given equation (x - 1) 2 / 9 + (y + 2) 2 / 16 = 1
a) Ellipse with center at (h , k) = (1 , -2) with a = 2 and b = 3.
b) its major axis is the line x = 1, and its minor is the line y = -2.
length of major axis = 2a = 8 , length of minor axis = 2b = 6
c) vertices at: (1 , -2 + 2) = (1 , 0) and (1 , -2 - 2) = (1 , -4)
d) c = √(a 2 - b 2) = √7
Foci at: (1 , -2+ √7) and (1 , -2 - √7)
Ques. Tangents are drawn from the point P(3, 4) to the ellipse x2/9 + y2/4 = 1 touching the ellipse at points A and B. Find their co-ordinates. (4 marks)
Solution: Given equation of ellipse is x2/9 + y2/4 = 1 …(1)
The equation of chord is
xx1/a2 + yy1/b2 = 1
3x/9 + 4y/4 = 1
=> x + 3y – 3 = 0
=> x = 3(-y + 1)
Now, \(\frac{9(-y+1)^2}{9} + \frac{y^2}{4}\)=1.
9(−y+1)29+y24=1
=> y2 – 2y + 1 + y2/4 = 1
=> 5y2 – 2y = 0
=> y(5y/4 – 2) = 0
=> y = 0 or y = 8/5
So, x = 3 – 3y
At y = 0 => x = 3
At y = 8/5 => x = -9/5
Therefore, A and B are (3, 0) and (-9/5, 8/5) respectively.
Ques. Find the equation of the ellipse whose axes are the axes of coordinates and which passes through the point (-3, 1) and has eccentricity √(1/5) (2 marks)
Solution: Given, eccentricity = √(1/5)
e2 = 1/5
we know, e2 = 1 – b2/a2
=> b2/a2 = 1 – 1/5 = 4/5
=> x2/5k + y2/4k = 1, this equation passes through the point (-3, 1)
=> (9/5) + (1/4) = k
Or k = 37/20
Therefore, required equation of ellipse is 4x2/37 + 5y2/37 = 1
Ques. Find the equation of ellipse if its foci and eccentricity are (±3,0) and 1/4 respectively. (2 marks)
Solution: Let the equation of the ellipse be \(\frac{x^2}{a^2}+ \frac{y^2}{b^2}\)=1. Then, coordinates of the foci are (±ae,0).
Therefore, ae = 4 ⇒ a = 12
We have b2 = a2(1–e2) ⇒ b2 =135
Thus, the equation of the ellipse is x2/144 + y2/135 = 1
Also read:
Comments