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 midpoint of the linesegment 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 midpoint of the linesegment 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 coordinates are (0, 0).
Solved examples to find the centre of an ellipse:
1. Find the coordinates 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 coordinates of the centre of the ellipse 3x\(^{2}\) + 2y\(^{2}\)  6 = 0 is (0, 0)
2. Find the coordinates 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 yaxes
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 coordinates of its centre are (α, β) i.e., (1,  5).
11 and 12 Grade Math
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