Parametric Equation of the Ellipse

We will learn in the simplest way how to find the parametric equations of the ellipse.

The circle described on the major axis of an ellipse as diameter is called its Auxiliary Circle.

If x2a2 + y2b2 = 1 is an ellipse, then its auxiliary circle is x2 + y2 = a2.

Let P (x, y) be any point on the equation of the ellipse be x2a2 + y2b2 = 1 ................... (i)

Now from P draw PM perpendicular to the major axis of the ellipse and produced MP cuts the auxiliary circle x2 + y2 = a2 at Q. Join the point C and Q. Again, let โˆ XCQ = ั„. The angle โˆ XCQ = ั„ is called the eccentric angle of the point P on the ellipse.

The major axis of the ellipse  x2a2 + y2b2 = 1 is AA' and its length = 2a. Clearly, the equation of the circle described on AA' as diameter is x2 + y2 = a2

Now, clearly we see that,

Since, CQ is the radius of the auxiliary circle x2 + y2 = a2

Therefore, CM = a cos ั„

or, x = a cos ั„.

Since the point P (x, y) lies on the ellipse x2a2 + y2b2 = 1, Therefore,

a2cos2ั„a2 + y2b2 = 1, (Since, x = a cos ั„)

โ‡’ y2b2 = 1 - cos2 ั„

โ‡’ y2 = b2(1 - cos2 ั„)

โ‡’ y2 = b2 sin2 ั„

โ‡’ y = b sin ั„

Hence, the co-ordinates of P are (a cos ั„, b sin ั„).

Therefore, for all values of ั„ the point P (a cos ั„, b sin ั„) always lies on the ellipse x2a2 + y2b2  = 1   

Thus, the co-ordinates of the point having eccentric angle ั„ can be written as (a cos ั„, b sin ั„). Here (a cos ั„, b sin ั„) are known as the parametric co-ordinates of the point P.

 

The equations x = a cos ั„, y = b sin ั„ taken together are called the parametric equations of the ellipse x2a2 + y2b2 = 1; where ั„ is parameter (ั„ is called the eccentric angle of the point P).

Note: The angle โˆ XCP is not the eccentric angle of the point P.


Solved example to find the parametric equations of an ellipse:

Find the equation to the auxiliary circle of the ellipse

4x2 + 9y2 - 24x - 36y + 36= 0.

Solution:

4x2 + 9y2 - 24x - 36y + 36 = 0

โ‡’ 4 (x2 - 6x + 9) + 9 (y2 - 4y + 4) = 36

โ‡’ (xโˆ’3)29 + (yโˆ’2)24 = 1 .............. (i)

Clearly, equation (i) represents an ellipse whose major axis is parallel to x-axis and centre is at (3, 2). Again, if the length of the major axis of the ellipse (i) be 2a then a2 = 9 โ‡’ a = 3.

Now, the circle described on the major axis of the ellipse (i) as diameter is its auxiliary circle.

Therefore, the centre of the auxiliary circle is at (3, 2) and its radius is 3.

Therefore, the required equation to the auxiliary circle of the ellipse (i) is

(x - 3)2 + ( y - 2 )2 = 32

โ‡’ x2 - 6x + 9 + y2 - 4y + 4 = 9

โ‡’ x2 + y2 - 6x - 4y + 4 = 0.




11 and 12 Grade Math 

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