Help plz 10 points and ill make urs best answer

for each ellipse, find the centers, each vertices and foci
1. X^2+2y^2-10x+8y+29=0
2. X^2 +4y^2-6x+24y+41=0

There's suppose to be 4 vertices

1)x^2+2y^2-10x+8y+29=0

x^2+2y^2-10x+8y =-29 complete squares

(x^2 -10x+25)+2(y^2+4y+4)=-29+ 25+2(4)

(x-5)^2+2(y+2)^2= 4 divide by 4

(x-5)^2/4 +(y+2)^2/2=1

The larger denominator is a^2, and the x part of the equation has the larger denominator, so this ellipse will be wider than tall (to parallel the x-axis). Also, a^2 = 4 and b^2 = 2, so the equation b^2 + c^2 = a^2 gives me 2 + c2 = 4, and c2 must equal 2.
The center is clearly at the point (h, k) = (5, -2).

a=2 and b=sqrt(2) then 2a=4 and 2b=2sqrt(2)

vertices (3,-2) and (7,-2)

(5,-2-sqrt(2)),(5,-2+sqrt(2))

Foci (5-sqrt(2),-2) and (5+sqrt(2), -2)


2) x^2 +4y^2-6x+24y+41=0

x^2 +4y^2-6x+24y= -41

(x^2-6x+9)+4(y^2+6y+9)=-41+9+4(9)

(x-3)^2+4(y+3)^2=36

(x-3)^2/36+(y+3)^2/9=1

The larger denominator is a^2, and the x part of the equation has the larger denominator, so this ellipse will be wider than tall (to parallel the x-axis). Also, a^2 =36 and b^2 = 9, so the equation b^2 + c^2 = a^2 gives me 9 + c2 = 36, and c2 must equal 27.
The center is clearly at the point (h, k) = (3, -3).

a=6 and b=3 then 2a=12 and 2b=6

vertices (-3,-3) and (9,-3)

(3,-6), (3,0)

Foci (3-sqrt(27),-3) and (3+sqrt(27), -3)

x^2 -10x +25 + 2*(y^2 +4y + 4) = 4
((x-5)^2 )/4 + ((y+2)^2)/2 = 1
center = 5,-2
vertices = (3,-2),(7,-2),(5,-2+root(2)),(5,-2-root(…
foci (5-root(2),-2), (5+root(2), -2)

similarly,
(x-3)^2/4 + (y-3)^2 = 1
a = 2, b =1
center = 3,3
vertices = (1,3) (5,3) (3,4),(3,2)
foci = (3+root(3),3) (3-root(3),3)
where f = Sqrt(a^2 - b^2) = sqrt(4 -1) = sqrt(3)