# Parametric Equation of an Ellipse Clearly, x = a cosθ, y = bsinθ satisfy the equation

$\large \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ; for all real values of θ

Hence (acos θ, b sinθ) is always a point on the ellipse

$\large \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ; for any value of θ

The point (a cosθ , b sinθ) is also called the point θ.

The angle θ is called the eccentric angle (0 ≤ θ < 2π ) of the point P(a cosθ , b sinθ) on the ellipse.

To figure out a point on the ellipse with eccentric angle θ we draw a circle with AA’ (the major axis) as the diameter.

This circle is called the auxiliary circle of the ellipse.

The equation of the circle is x2 + y2 = a2 We draw ∠ACQ= θ . Then Q ≡ (a cosθ, a sinθ). Draw QM as perpendicular to AA’ cutting the ellipse at P. The x-co-ordinate of P = CM = a cosθ

=> y – coordinate of P is b sinθ

=> P ≡ (a cosθ, b sinθ).

Illustration : Find the centre, the lengths of the axes and the eccentricity of the ellipse
2x2 + 3y2 − 4x − 12y + 13 = 0

Solution: The given equation can be written as

2(x − 1)2 + 3(y − 2)2 = 1

$\large \frac{(x-1)^2}{1/2} + \frac{(y-2)^2}{1/3} = 1$

=> The centre of the ellipse is (1 , 2).

The major axis = 2/√2 = √2

The minor axis = 2/√3

=> e2 = 1 − (b2/a2)

e = 1 − 2/3

= 1/3

=> e = 1/√3

Illustration : Find the equation of the ellipse whose foci are (2, 3), (–2, 3) and whose semi minor axis is of length √5

Solution:

Here S is (2, 3) , S’ is (–2 , 3) and b = √5

⇒ SS’ = 4 = 2ae

⇒ ae = 2

But b2 = a2 (1 – e2)

⇒ 5 = a2 – 4 ⇒ a = 3.

If P(x, y) is any point on the ellipse,

then SP + S’P = 2a.

$\large \sqrt{(x-2)^2 + (y-3)^2 } + \sqrt{(x+2)^2 + (y-3)^2} = 6$ . . . (1)

But [(x – 2)2 + (y – 3)2] – [(x + 2)2 + (y – 3)2] = – 8x . . . (2)

$\large \sqrt{(x-2)^2 + (y-3)^2 } – \sqrt{(x+2)^2 + (y-3)^2} = -\frac{4 x}{3}$ . . . (3)

(On dividing (2) by (1)). And then adding (1) and (3), we get

$\large \sqrt{(x-2)^2 + (y-3)^2 } = 3-\frac{2 x}{3}$

⇒ 5x2 + 9y2 – 54y + 36 = 0.

Alternative:

Here S ≡ (2, 3) ; S’ ≡ (–2, 3) and b = √5

SS’ = 2ae = 4

⇒ ae = 2

b2 = a2 (1 – e2)

⇒ 5 = a2 – 4

⇒ a = 3

Hence the equation to major axis is y = 3

Centre of ellipse is midpoint of SS’ i.e. (0, 3)

∴ Equation to ellipse is

$\large \frac{x^2}{a^2} + \frac{(y-3)^2}{b^2} = 1$

or , $\large \frac{x^2}{9} + \frac{(y-3)^2}{5} = 1$

Illustration : Find the equation of the ellipse having centre at (1, 2), one focus at (6, 2) and passing through the point (4, 6).

Solution:

With centre at (1 , 2) , the equation of the ellipse is

$\large \frac{(x-1)^2}{a^2} + \frac{(y-2)^2}{b^2} = 1$

It passes through the point (4, 6)

$\large \frac{9}{a^2} + \frac{16}{b^2} = 1$ . . . .(1)

Distance between the focus and the centre

= (6 – 1) = 5 = ae

⇒ b2 = a2 – a2e2

= a2 – 25 . . . .(2)

Solving for a2 and b2 from the equations (1) and (2) , we get a2 = 45 and b2 = 20.

Hence the equation of the ellipse is

$\large \frac{(x-1)^2}{45} + \frac{(y-2)^2}{20} = 1$

Alternative:

Centre is C (1 , 2); One focus is S(6 , 2)

Second focus is S’ ≡ (- 4 , 2)

⇒ The major axis is along y = 2

Also CS = ae = 5 . . . . (1)

and P(4, 6) is a point on the ellipse.

Hence SP + S’P = 2a yields

2√5 + 4√5 = 2a

⇒ a = 3√5

⇒ e = √5/3

and b2 = a2(1 – e2)

= 45(1−5/9)

= 20

Hence the equation to the ellipse is

$\large \frac{(x-1)^2}{45} + \frac{(y-2)^2}{20} = 1$

Illustration : Consider the ellipse x2 + 3y2 = 6 and a point P on it in the first quadrant at a distance of 2 units from the centre. Find the eccentric angle of P.

Solution Equation of ellipse is x2 + 3y2 = 6

Equation of auxiliary circle is x2 + y2 = 6

Since P ≡ (x1, y1) & Q ≡ (x1, y2) lie on the ellipse and the circle respectively

we have,

x12 + 3y12 = 6 …..(1)

x12 + y22 = 6 …..(2)

∴ 3y12 – y22 = 0

⇒ y2 = √3y1

Again OP = 2

⇒ x12+ y12= 4 ….(3)

By (1) –(3) , we get ,

2y12 = 2

⇒ y12 = 1 ⇒ y1 = 1

[∴ P is in the first quadrant]

∴ y2 = √3

Putting y1 in (1), we get, x12 = 3

x1 = √3

∴ Eccentric angle of P

= θ = tan-1(y2/x1)

= tan-1(√3/√3)

= π/4

Exercise :

(i) Obtain the equation to an ellipse whose focus is the point (–1, 1), whose directrix is the line x – y + 3 = 0 and whose eccentricity is 1/2.
(ii) Find the centre, the length of axes, the eccentricity and the foci of the ellipse
12 x2 + 4 y2 + 24x – 16y + 25 = 0.

(iii) Find the eccentricity of an ellipse, if its latus rectum is equal to one half of its major axis.

(iv) Find the equation to the ellipse whose one vertex is (3, 1), the nearer focus is (1, 1) and the eccentricity is 2/3.

(v) Find the latus rectum, eccentricity and foci of the curve 4x2 + 9y2– 8x– 36y + 4 = 0