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).

● The Ellipse

• Definition of Ellipse
• Standard Equation of an Ellipse
• Two Foci and Two Directrices of the Ellipse
• Vertex of the Ellipse
• Centre of the Ellipse
• Major and Minor Axes of the Ellipse
• Latus Rectum of the Ellipse
• Position of a Point with respect to the Ellipse
• Ellipse Formulae
• Focal Distance of a Point on the Ellipse
• Problems on Ellipse

11 and 12 Grade Math 

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