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We will discuss about the centre of the ellipse along with the examples.
The centre of a conic section is a point which bisects every chord passing through it.
Definition of the centre of the ellipse:
The mid-point of the line-segment joining the vertices of an ellipse is called its centre.
Suppose the equation of the ellipse be x2a2 + y2b2 = 1 then, from the above figure we observe that C is the mid-point of the line-segment AA', where A and A' are the two vertices. In case of the ellipse x2a2 + y2b2 = 1, every chord is bisected at C (0, 0).
Therefore, C is the centre of the ellipse and its co-ordinates are (0, 0).
Solved examples to find the centre of an ellipse:
1. Find the co-ordinates of the centre of the ellipse 3x2 + 2y2 - 6 = 0.
Solution:
The given equation of the ellipse is 3x2 + 2y2 - 6 = 0.
Now form the above equation we get,
3x2 + 2y2 - 6 = 0
⇒ 3x2 + 2y2 = 6
Now dividing both sides by 6, we get
x22 + y23 = 1 ………….. (i)
This equation is of the form x2a2 + y2b2 = 1 (a2 > b2).
Clearly, the centre of the ellipse (1) is at the origin.
Therefore, the co-ordinates of the centre of the ellipse 3x2 + 2y2 - 6 = 0 is (0, 0)
2. Find the co-ordinates of the centre the ellipse 5x2 + 9y2 - 10x + 90y + 185 = 0.
Solution:
The given equation of the ellipse is 5x2 + 9y2 - 10x + 90y + 185 = 0.
Now form the above equation we get,
5x2 + 9y2 - 10x + 90y + 185 = 0
⇒ 5x2 - 10x + 5 + 9y2 + 90y + 225 + 185 - 5 - 225 = 0
⇒ 5(x2 - 2x + 1) + 9(y2 + 10y + 25) = 45
(x−1)29 + (y+5)25 = 1
We
know that the equation of the ellipse having centre at (α, β) and major and minor axes parallel to x and y-axes
respectively is, (x−α)2a2 + (y−β)2b2 = 1.
Now, comparing equation (x−1)29 + (y+5)25 = 1 with equation (x−α)2a2 + (y−β)2b2 = 1 we get,
α = 1, β = - 5, a2 = 9 ⇒ a = 3 and b2 = 5 ⇒ b = √5.
Therefore, the co-ordinates of its centre are (α, β) i.e., (1, - 5).
● The Ellipse
11 and 12 Grade Math
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