Asked by Jaypee
X² + 4y² - 6x + 8y + 9 = 0
Answers
Answered by
Bosniam
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
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
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