Question

The x - coordinates of the foci of the ellipse x^2 – 2x + 2y^2 = 3 are:
Options :

Answer 1

Given :

Equation of the given ellipse :

$x^2 -2x + 2y^2 =3$

To Find :

What are the x -coordinates of the foci of this ellipse ?

Solution :

On rearranging the given equation of ellipse we can obtain the equation in general form as :

$x^2 - 2x +1 + 2y^2 -1 =3$  ( adding and subtracting 1  )

Or,$(x-1)^2 + 2y^2 = 4$

Or,$\frac{(x-1)^2}{4} + \frac{2y^2}{4}=1$

Or, $\frac{(x-1)^2}{2^2} + \frac{y^2}{(\sqrt2)^2} = 1$

So the centre of this ellipse is at (1,0). And also this ellipse is a horizontal ellipse as value of a>b .

Now we know that the distance between centre and foci of an ellipse is 'ae' . So to find 'ae' we have :

∴$(ae)^2 = a^2 -b^2$

          = $2^2 - (\sqrt2)^2$

          =$4 - 2$

          = 2

Or, $ae = \sqrt2$

Now we can easily find the coordinates of the foci of this ellipse by adding $\sqrt2$ to the center point. So we will add $\sqrt2$ to the x coordinate in the center coordinates to the left and right i.e. to  (2,0) .

So the coordinates of the foci of this ellipse  will be $(2-\sqrt2,0)$ and $(2+\sqrt2,0)$

Hence , the x coordinates of this ellipse is $2-\sqrt2$ and $2+\sqrt2$ .