Put into standard form and identfy the vertcies, co-verticies, and foci?

Question

9x^2 + 4y^2 -18x + 24y +9 = 0

Answer 1

 9x^2 + 4y^2 - 18x + 24y + 9  = 0

 9x^2 + 4y^2 - 18x + 24 y + 9 +36  =  36    (Add 36 to each side)

9x^2 - 18 x + 9 + 4y^2 + 24y + 36  = 36

( 3x - 3 )^2 + ( 2y + 6 )^2  = 36

Divide  each  side  by  36

( 3x - 3 )^2 / 36  + ( 2y + 6 )^2 / 36 = 1

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

( x - 1)^2 / 12  + ( y + 3 )^2/ 18  = 1

[ x - 1]^2 / ( 12 ) + [ y - ( -3 ) ]^2 / ( 18 )  = 1

The  conic is ellipse.

 

Comments

Standard form of  is .

Answer 2

The conic equation

To identify the conic section

General form of a conic equation in the form

Both variables are squared and have the same sign, but they aren't multiplied by the same number, so this is an ellipse.

is ellipse.

To find the ellipse in standard form.

To change the expressions (x ²- 2x) and (y ² + 6y) into a perfect square trinomial,

add (half the x coefficient)² and add (half the y coefficient)² to each side of the equation

The standard form for an ellipse is in a form = 1, So divide both sides of equation by 36 to set it equal to 1.

Compare it to standard form of ellipse

a ² > b ²

If the larger denominator is under the "y " term, then the ellipse is vertical.

Center (h, k ) = (1,-3)

a  = length of semi-major axis = 3

b  = length of semi-minor axis = 2

Vertices: (h, k + a ), (h, k - a )

 = (1, -3+3) , (1, -3-3)

Vertices : (1, 0), (1, -6)

Co-vertices: (h + b, k), (h - b, k) [endpoints of the minor axis]

= (1+2, -3), (1-2, -3)

Co-vertices : (3, -3) ,(-1, -3)

c  is the distance from the center to each focus.

foci: (h, k + c ), (h, k - c )

Foci : .