Table of Contents
Last modified on August 3rd, 2023
An ellipse is a closed curved plane formed by a point moving so that the sum of its distance from the two fixed or focal points is always constant. It is formed around two focal points, and these points act as its collective center.
We can also define ellipse as the shape formed when a straight plane intersects a right circular cone at an angle greater than zero degrees.
The flatness or roundness of an ellipse is determined by a property termed eccentricity. It is represented by ‘e’.
Ellipses can have only two orientations based on planes (x-axis or y-axis) the foci lie on.
Horizontal ellipse
Ellipse having the major axis on the x-axis.
Vertical Ellipse
Ellipse having the major axis on the y-axis.
The eccentricity of an ellipse measures how flat or round an ellipse is.
The more flat an ellipse is, the greater is its eccentricity, and the more round it is, its eccentricity is closer to zero. Eccentricity as 1 is a straight line, and zero will be a perfect circle. It is less than 1. (e<1).
The formula for the eccentricity of an ellipse is given below:
Here we will derive the above formula of the eccentricity of the ellipse.
O is the center at the origin. F1 (-c,0) and F2 (c,0) are the foci.
A & B are the vertices. AB = major axis = 2a
As we know,
c= 1/2 distance between the 2 focal points, and a= semi-major axis
P (0,b) is a point on ellipse.
Let PO=b=semi-minor axis, PF1 + PF2 = 2a, OF2=c,
As we know, ∠POB=∠POA=90°, therefore, ∠POF2 = 90°
Applying Pythagoras theorem for right-angled △POF2,
∴ PF2 =√(b2 + c2), and also PF2 = a
∴ c= √(a2-b2)
Putting the value of c in the formula of eccentricity, we get,
e = √(1-b2/a2)
= √{( a2-b2)/a2}
= √(c2/a2) =c/a
∴ e=c/a
Thus we get the standard simplified formula of the eccentricity of the ellipse.
Note: If a=b, (like in the case of a circle)
then c =√(a2-b2) = 0
∴ e = 0 (eccentricity of circle is zero)
∴ 0≤e>1
Let us solve an example involving the eccentricity of the ellipse.
Find the eccentricity of an ellipse with a major axis 12 and the distance between the focus and center is 5.
As we know,
Eccentricity (e) = c/a, here, c = 5 units, a = 6 units
= 5/6 = 0.833 units
The standard form of equation of an ellipse is x2/a2 + y2/b2 = 1, where a = semi-major axis, b = semi-minor axis.
Let us derive the standard equation of an ellipse centered at the origin
The equation of ellipse focuses on deriving the relationships between the semi-major axis, semi-minor axis, and the focus-center distance.
Our aim is to find the relationships of a, b, and c.
Length of major axis = 2a.
Length of minor axis = 2b.
Distance between the foci = 2c.
Take another point Q on one end of the minor axis.
Find the sum of the distances of Q from foci F1 and F2
QF1+ QF2 = √(b2+c2)+ √(b2+c2)
QF1+ QF2 = 2 [√(b2+c2)]
As Q lie on the ellipse. As per the definition, sum of distances between any point on ellipse with the foci is constant. (2a according to the diagrams)
2√(b2+c2)= 2a
√(b2+c2)= a
Squaring both sides:
b2 + c2 = a2
c2 = a2 – b2
SF1 + SF2 = 2a
Now applying the distance formula, we get
=>√(x+c)2+y2 + √(x−c)2+y2 = 2a
√[(x+c)2+y2]= 2a – √[(x−c)2+y2]
Squaring both sides
(x + c)2 + y2 = 4a2 + (x – c)2 + y2 – 4a√(x−c)2+y2
x2 + c2 + 2cx + y2
= 4a2 + x2 + c2 – 2cx + y2 – 4a√[(x−c)2+y2]
4cx – 4a2 = – 4a√[(x−c)2+y2]
a2 – cx = a√[(x−c)2+y2]
Squaring both sides and simplifying, we have
x2/a2+y2 /(c2−a2)=1
As we have c2 = a2 – b2 we can substitute this in the above equation.
We can substitute this in the above equation
x2/a2+y2/b2=1
This is the standard equation of an ellipse centered at the origin.
Here, P (x,y) is a point on the ellipse, major axis = 2a, minor axis = 2b
Let us solve few examples involving equations of an ellipse
Find the equation of the ellipse with a major axis on the X-axis and having coordinates (2, -2) and (-1, 3)
As we know, the standard equation of an ellipse is given by
x2 /a2+y2/b2 = 1, here, points (2, 2) and (-1, 3) lie on the ellipse,
Now, there can be two possible situations:
Case 1: (2)2/a2 + (2)2/b2 =1, when it has coordinates (2, 2)
Case 2: (-1)2/a2 + (3)2/b2 =1, when it has coordinates (-1, 3)
Comparing with the standard equation, we get
a2 = 35/8
b2 =35/3
Putting values of a and b on the equation of an ellipse, we get
x2 ÷ (35/8) + y2 ÷ (35/3) =1
8x2 + 3y2=35
Find the values of a (major axis) and b (minor axis) from the equation 4x2 + 25y2 =100.
4x2 + 25y2 =100 (Given)
Dividing by 100 at both sides
=> x2/25 + y2 /4 = 1
Comparing this equation with equation of an ellipse, x2/a2 + y2 /b2=1
=>a2 = 25 => a=5
=> b2 = 4 => b=2
Thus the value of the major axis (2a) = 10 and minor axis (2b) = 4
The formula to find the area of an ellipse is given below:
Now, let us solve a problem involving the area of an ellipse to clear your concept better.
Find the area of an ellipse whose major axis is 6 units and the minor axis is 2 units in length.
Area (A) = πab, here a = 3, b = 1, π = 3.141 = 22/7
= π x 3 x 1
= π x3
= 9.424 sq units
Thus the area of the ellipse is 9.424 sq units
We generally consider the half of any two-dimensional shape when talking about the area. We already have found out the area of an ellipse in the above example. Now let’s talk about the half of an ellipse.
Semi-ellipse is one-half of an ellipse. When an ellipse is cut, it forms a curved side (arc) and a straight side. The straight side is called the height or Rise. It is denoted as H of the semi-ellipse.
Since the area of an ellipse is πab.
Therefore,
The formula of the area of a semi-ellipse is 1/2(πab) sq. units
Let us solve an example.
Find the area of a semi-ellipse whose major axis is 8 units and the minor axis is 2 units in length.
As we know,
Area (A) = 1/2(πab), here a = 4, b = 1
= 1/2(π x 4 x 1) sq. units
= 6.283 sq. units.
The formula to find the perimeter of an ellipse is given below:
Now, let us solve a problem involving the perimeter of an ellipse to clear your concept better.
Find the perimeter of an ellipse whose major axis is 6 units and the minor axis is 2 units in length.
Perimeter (P) = 2π x √(a2 + b2)/2, here a = 3, b = 1, π = 3.141 = 22/7
=2π [√(32 + 12)/2)
= 2π√5
=14.049 units
Thus the perimeter of the ellipse is 14.049 units
The two-dimensional shape of a rugby ball.
Whispering galleries. Two people standing under a semi-elliptical roof can whisper and communicate with each other. Grand Central Terminal in New York is one of the real examples of such galleries.
Lithotripter, a medical device, applies the principle of an ellipse. It helps to break the stones in kidneys by shockwaves applied from one focal point in the ellipsoidal reflector in the device, whereas; the kidney is placed at the other focal point.
The orbital movement of the planets in our solar system is also in an elliptical pattern.
Last modified on August 3rd, 2023