Centre of the Ellipse

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 \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 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 \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 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 3x\(^{2}\) + 2y\(^{2}\) - 6 = 0.

Solution:

The given equation of the ellipse is 3x\(^{2}\) + 2y\(^{2}\) - 6 = 0.

Now form the above equation we get,

3x\(^{2}\) + 2y\(^{2}\) - 6 = 0

3x\(^{2}\) + 2y\(^{2}\) = 6

Now dividing both sides by 6, we get

\(\frac{x^{2}}{2}\) + \(\frac{y^{2}}{3}\) = 1 ………….. (i)

This equation is of the form \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 (a\(^{2}\) > b\(^{2}\)).

Clearly, the centre of the ellipse (1) is at the origin.

Therefore, the co-ordinates of the centre of the ellipse 3x\(^{2}\) + 2y\(^{2}\) - 6 = 0 is (0, 0)

 

2. Find the co-ordinates of the centre the ellipse 5x\(^{2}\) + 9y\(^{2}\) - 10x + 90y + 185 = 0.

Solution:    

The given equation of the ellipse is 5x\(^{2}\) + 9y\(^{2}\) - 10x + 90y + 185 = 0.

Now form the above equation we get,

5x\(^{2}\) + 9y\(^{2}\) - 10x + 90y + 185 = 0

5x\(^{2}\) - 10x + 5 + 9y\(^{2}\) + 90y + 225 + 185  - 5 - 225 = 0

5(x\(^{2}\) - 2x + 1) + 9(y\(^{2}\) + 10y + 25) =  45

\(\frac{(x - 1)^{2}}{9}\) + \(\frac{(y + 5)^{2}}{5}\) = 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, \(\frac{(x - α)^{2}}{a^{2}}\) + \(\frac{(y - β)^{2}}{b^{2}}\) = 1.

Now, comparing equation \(\frac{(x - 1)^{2}}{9}\) + \(\frac{(y + 5)^{2}}{5}\) = 1 with equation \(\frac{(x - α)^{2}}{a^{2}}\) + \(\frac{(y - β)^{2}}{b^{2}}\) = 1 we get,

α = 1, β = - 5, a\(^{2}\) = 9 a = 3 and b\(^{2}\) = 5 b = √5.

Therefore, the co-ordinates of its centre are (α, β) i.e., (1, - 5).





11 and 12 Grade Math 

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