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

Answer:(5, -2)Step-by-step explanation:The equation of the ellipse is given by:               [tex]$ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 $[/tex]where [tex]$ (h, k)  $[/tex] is the center of the ellipse.The given equation of the ellipse is: x² + 2y² - 10x + 8y + 25 = 0.We have to reduce to the standard form so that we can compare and determine the center of the ellipse.Subtracting [tex]25[/tex] from the equation on both sides, we get: x² - 10x + 2y² + 8y + 25 - 25 = - 25[tex]$ \implies x^2 - 10x + 2y^2 + 8y = - 25 $[/tex][tex]$ \implies x^2 - 10x + 2(y^2 + 4y) = -25 $[/tex]The next step would be complete the squares. Let us complete [tex]x[/tex] first.[tex]x^2 - 10x = x^2 - 5(2)x \\= x^2 - 5(2)x - 25 + 25  \\= (x - 5)^2 + 25[/tex]Now for [tex]y[/tex].[tex]2(y^2 + 4y) =2\{ y^2 + 2(2)y \} \\\\= 2\{y^2 + 2(2)y + 4 - 4 \} \\\\= 2\{(y + 2y)^2 - 4\} \\\\= 2\{(y + 2)^2\} - 8[/tex]Therefore the equation becomes: (x - 5)² + 2(y + 2)² = -25 + 25 + 8⇒ (x - 5)² + 2(y - 2)² = 8Since the [tex]R.H.S.[/tex] needs 1, we divide the entire equation by 8.⇒ [tex]$ \frac{(x - 5)^2}{8} + 2\frac{(y - 2)^2}{8} = 1 $[/tex][tex]$ \implies \frac{(x - 5)^2}{8} + \frac{(y + 2)^}{4} = 1 $[/tex]Comparing with standard form we get:[tex]$ (x - h)^2 = (x - 5)^2 $[/tex] and [tex]$ (y - k)^2 = (y + 2)^2 $[/tex]⇒ (h,k) = (5,-2)∴ The center of the ellipse is (5,-2).