X² + 4y² - 6x + 8y + 9 = 0

Answer 1

That is not calculus.

x ² + 4 y² - 6 x + 8 y + 9 = 0

Re-group:

Subtract 9 to both sides.

x ² + 4 y² - 6 x + 8 y = - 9

( x² - 6 x ) + ( 4 y² + 8 y ) = - 9

( x² - 6 x ) + 4 ( y² + 2 y ) = - 9

Complete the squares:

( x² - 6 x + 3² ) + 4 ( y² + 2 y + 1² ) = - 9 + 3² + 4 ∙ 1² = - 9 + 9 + 4

( x - 3 )² + 4 ( y + 1 )² = 4

Divide both sides by 4

( x - 3 )² / 4 + ( y + 1 )² = 1

( x - 3 )² / 4 + ( y + 1 )² / 1 = 1

( x - 3 )² / 2² + ( y + 1 )² / 1² = 1

Now:

( x - h )² / a² + ( y - k )² / b² = 1

is the ellipse standard equation where

h and k are x and y coordinate of center

a and b are the semi-major and semi-minor axes

x ² + 4 y² - 6 x + 8 y + 9 = 0

which is the same as

( x - 3 )² / 2² + ( y + 1 )² / 1² = 1

is ellipse with:

center

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

semi-major axes a = 2

semi-minor axes b = 1