Question

Help with putting conic equation into vertex form

Answer 1

x2 + 9y2 + 10x − 18y + 25 = 0

Answer 2

I know this is an ellipse

Answer 3

And it is a horizontal ellipses as well

Answer 4

complete the square twice
it is a real pain

Answer 5

\[\huge\rm (x^2+10x)+(9y^2-8y)=-25\]
complete the square
left side

Answer 6

\[x^2 + 9y^2 + 10x − 18y + 25 = 0 \\
x^2-10x+9y^2-18y=25\] is a start

Answer 7

(x^2+10x) + (9y^2-18y)=-25

Answer 8

x2 + 9y2 + 10x − 18y + 25 = 0

$$x^2+9y^2+10x-18y+25=(x^2+10x+25)+9(y^2+2y+1-1)\\\quad=(x+5)^2+9(y+1)^2-9$$

Answer 9

factor out the 9 of the y term
\[x^2+10x+9(y^2-2y)=-25\]

Answer 10

(x+5)^2+9(y^2-2y)^2=-25

Answer 11

er, that should be \(y^2-2y+1-1\) and then \((y-1)^2\)

Answer 12

b/2 = 10/2
take square of 5 add to the right side

Answer 13

\[(x+5)^2+\overbrace{9(y^2-2y)^2}^{\text{mistake here}}=-25\]

Answer 14

\[(x-5)^2+9(y-1)^2=-25+5^2+9\]

Answer 15

so we are left with (x-5)^2 +9(y-1)^2=9?

Answer 16

then divide by 9 to put in standard form

Answer 17

i made a typo \(-5\) should be \(+5\) of course

Answer 18

it should equal to 1

Answer 19

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

Answer 20

nope

Answer 21

divide ALL by 9