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 \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 is an ellipse, then its auxiliary circle is x\(^{2}\) + y\(^{2}\) = a\(^{2}\).
Let P (x, y) be any point on the equation of the ellipse be \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 ................... (i)
Now from P draw PM perpendicular to the major axis of the ellipse and produced MP cuts the auxiliary circle x\(^{2}\) + y\(^{2}\) = a\(^{2}\) at Q. Join the point C and Q. Again, let ∠XCQ = ф. The angle ∠XCQ = ф is called the eccentric angle of the point P on the ellipse.
The major axis of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 is AA' and its length = 2a. Clearly, the equation of the circle described on AA' as diameter is x\(^{2}\) + y\(^{2}\) = a\(^{2}\)
Now, clearly we see that,
Since, CQ is the radius of the auxiliary circle x\(^{2}\) + y\(^{2}\) = a\(^{2}\)
Therefore, CM = a cos ф
or, x = a cos ф.
Since the point P (x, y) lies on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1, Therefore,
\(\frac{a^{2}cos^{2} ф}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1, (Since, x = a cos ф)
⇒ \(\frac{y^{2}}{b^{2}}\) = 1  cos\(^{2}\) ф
⇒ y\(^{2}\) = b\(^{2}\)(1  cos\(^{2}\) ф)
⇒ y\(^{2}\) = b\(^{2}\) sin\(^{2}\) ф
⇒ y = b sin ф
Hence, the coordinates of P are (a cos ф, b sin ф).
Therefore, for all values of ф the point P (a cos ф, b sin ф) always lies on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1
Thus, the coordinates of the point having eccentric angle ф can be written as (a cos ф, b sin ф). Here (a cos ф, b sin ф) are known as the parametric coordinates of the point P.
The equations x = a cos ф, y = b sin ф taken together are called the parametric equations of the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1; where ф is parameter (ф is called the eccentric angle of the point P).
Note: The angle ∠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
4x\(^{2}\) + 9y\(^{2}\)  24x  36y + 36= 0.
Solution:
4x\(^{2}\) + 9y\(^{2}\)  24x  36y + 36 = 0
⇒ 4 (x\(^{2}\)  6x + 9) + 9 (y\(^{2}\)  4y + 4) = 36
⇒ \(\frac{(x  3)^{2}}{9}\) + \(\frac{(y  2)^{2}}{4}\) = 1 .............. (i)
Clearly, equation (i) represents an ellipse whose major axis is parallel to xaxis and centre is at (3, 2). Again, if the length of the major axis of the ellipse (i) be 2a then a\(^{2}\) = 9 ⇒ 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 )\(^{2}\) = 3\(^{2}\)
⇒ x\(^{2}\)  6x + 9 + y\(^{2}\)  4y + 4 = 9
⇒ x\(^{2}\) + y\(^{2}\)  6x  4y + 4 = 0.
`11 and 12 Grade Math
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