what is standard form of the ellipse equation: 4X^2 + 8X + 9Y^2 -54Y + 49 = 0

Question

I know to complete the square for each the X and Y terms, but can't deal with the -49 brought over to the right side of the equation

Answer 1

Given equation is 4x^2+8x+9y^2-54y+49 = 0

4x^2+8x+9y^2-54y+85-45 = 0

4x^2+8x+9y^2-54y+81+4 = 45

4x^2+8x+4+9y^2-54y+81 = 45

4(x^2+2x+1)+9(y^2-6y+9) = 45

4(x+2)2+9(y-3)^2 = 45

Divide to each side by 45

4(x+2)^2/45+9(y-3)^2/45 = 45/45

(x+2)^2/9+(y-3)^2/4 = 1

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

If the large denominator is under the x term, then the ellpse is horizontal ellipse.

Form is(x-h)^2/a^2+(y-k)^2/b^2 = 1, a > b.

Standard form of given ellipse is (x+2)^2/3^2+(y-3)^2/2^2 = 1.

Comments

Standard form of the  ellipse equation is .

Answer 2

The ellipse equation

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)²to 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 "x" term, then the ellipse is horizontal.

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

a  = length of semi-major axis = 3

b  = length of semi-minor axis = 2.

Standard form of the  ellipse equation is .