Question

Identify the center of the ellipse with equation x^2+2y^2-10x+8y+25=0

Answer 1

(5, -2)

Explanation:

The equation of the ellipse is given by:

`$ ((x - h)^2)/(a^2) + ((y - k)^2)/(b^2) = 1 $`

where
`$ (h, k)  $` is the center of the ellipse.

The given equation of the ellipse is:

x² + 2y² - 10x + 8y + 25 = 0.

We have to reduce to the standard form so that we can compare and determine the center of the ellipse.

Subtracting
`25` from the equation on both sides, we get:

x² - 10x + 2y² + 8y + 25 - 25 = - 25

`$ \implies x^2 - 10x + 2y^2 + 8y = - 25 $`

`$ \implies x^2 - 10x + 2(y^2 + 4y) = -25 $`

The next step would be complete the squares. Let us complete
`x` first.

`x^2 - 10x = x^2 - 5(2)x \\= x^2 - 5(2)x - 25 + 25  \\= (x - 5)^2 + 25`

Now for
`y`.

`2(y^2 + 4y) =2\{ y^2 + 2(2)y \} \\\\= 2\{y^2 + 2(2)y + 4 - 4 \} \\\\= 2\{(y + 2y)^2 - 4\} \\\\= 2\{(y + 2)^2\} - 8`

Therefore the equation becomes:

(x - 5)² + 2(y + 2)² = -25 + 25 + 8

⇒ (x - 5)² + 2(y - 2)² = 8

Since the
`R.H.S.` needs 1, we divide the entire equation by 8.

⇒
`$ ((x - 5)^2)/(8) + 2((y - 2)^2)/(8) = 1 $`

`$ \implies ((x - 5)^2)/(8) + ((y + 2)^)/(4) = 1 $`

Comparing with standard form we get:

`$ (x - h)^2 = (x - 5)^2 $` and
`$ (y - k)^2 = (y + 2)^2 $`

⇒ (h,k) = (5,-2)

∴ The center of the ellipse is (5,-2).