Conic Sections

Equations of Ellipse

The set of all points in a plane, the sum of whose distances from two fixed points in the plane is constant is an ellipse. These fixed points (two) are the foci of the ellipse (Fig. 1). When a line segment is drawn joining the two focus points, then the mid-point of this line is the centre of the ellipse.

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This line (joining the two foci) is called the major axis and a line drawn through the centre and perpendicular to the major axis is the minor axis. The endpoints are called vertices of the ellipse (Fig. 2).


Also, we denote

  • The length of the major axis by ‘2a’
  • Length of the minor axis by ‘2b’
  • The distance between the foci by ‘2c’.

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  • The length of the semi-major axis is ‘a’
  • Semi-minor axis is ‘b’ and
  • The distance of focus from the centre is ‘c’


The Relationship Between ‘a’, ‘b’, and ‘c’

Take a look at the following diagram:


As shown, take a point P at one end of the major axis. Hence, the sum of the distances between the point P and the foci is,

F1P + F2P = F1O + OP + F2P = c + a + (a – c) = 2a.

Next, take a point Q at one end of the minor axis. Now, the sum of the distances between the point Q and the foci is,

F1Q + F2Q = √ (b2 + c2) + √ (b2 + c2) = 2√ (b2 + c2)

We know that both points P and Q lie on the ellipse. Hence, by definition we have

2√ (b2 + c2) = 2a
Or, √ (b2 + c2) = a
i.e. a2 = b2 + c2 or c2 = a2 – b2

Special Cases

In the equation, c2 = a2 – b2, if we keep a fixed and vary the value of ‘c’ from ‘0-to-a’, then the resulting ellipses will vary in shape.

  • Case-I c = 0: When c = 0, both the foci merge together at the centre of the figure. Also, a2 becomes equal to b2, i.e. a = b. Hence, the ellipse becomes a circle.
  • Case-II c = a: When c = a, b = 0. Hence, the ellipse reduces to a line joining the two points F1 and F2.
  • Eccentricity: It is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse. It is denoted by ‘e’. Therefore, e = c/a.

Standard Equations of an Ellipse

When the centre of the ellipse is at the origin and the foci are on the x or y-axis, then the equation of the ellipse is the simplest. Here are two such possible orientations:


Of these, let’s derive the equation for the ellipse shown in Fig.5 (a) with the foci on the x-axis. Let F1 and F2 be the foci and O be the mid-point of the line segment F1F2. Also, let O be the origin and the line from O through F2 be the positive x-axis and that through F1 as the negative x-axis.

Further, let the line drawn through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c, 0) and F2 be (c, 0) as shown in Fig.5 (a) above.

Derivation of the Equation

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 the 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)
We square both sides again and simplify it further to 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)

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)}

Simplifying the equation and replacing b2 with a2 – c2, 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, lies on the ellipse. Also, the equation of an ellipse with 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, we can say that the ellipse lies between the lines x = – a and x = a and touches these lines. Similarly, it can lie between the lines y = – b and y = b and touch those 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.


  • An ellipse is symmetric with respect to both the 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.

Latus Rectum

Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose endpoints lie on the ellipse as shown below.


Let’s find the length of the latus rectum of the ellipse x2/a2 + y2/b2 = 1 shown above. Let the length of AF2 be l. Therefore, the coordinates of A are (c, l).

∴ x2/a2 + y2/b2 = c2/a2 + l2/b2 = 1

Now, in ‘Eccentricity’ we learned that e = c/a or c = ae. Substituting the values, we get

(ae)2/a2 + l2/b2 = 1
Or, e2 + l2/b2 = 1
l2 = b2(1 – e2)

Now, we know that e = c/a
∴ e2 = c2/a2 = (a2 – b2)/a2 = 1 – b2/a2 … (from the relationship between a, b, and c)

Hence, we have
l2 = b2(1 – e2) = b2(1 – {1 – b2/a2}) = b2(1 – 1 + b2/a2) = b4/a2
Or, l = b2/a

Since the ellipse is symmetric with respect to y-axis,

AF2 = F2B

So, the length of the latus rectum is = 2b2/a.

Solved Examples for You

Question: Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse 9x2 + 4y2 = 36.

Solution: The equation given is, 9x2 + 4y2 = 36. Dividing both sides by 36, we get

x2/4 + y2/9 = 1

Observe that the denominator of y2 is larger than that of x2. Hence, the major axis is along the y-axis. Now, comparing it with the standard equation, we get, 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


  • 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
  • Eccentricity = √5/3
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