Question 118267
The standard form for the equation of an ellipse with center at (h, k) and major axis parallel to the x-axis is: (a is the semi-major axis and b is the semiminor axis)
{{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1}}} or, if the major axis is parallel to the y-axis:
{{{(x-h)^2/b^2 + (y-k)^2/a^2 = 1}}}
Starting with the given equation:
{{{3x^2+4y^2+8y = 8}}} Group the terms as shown:
{{{(3x^2)+(4y^2+8y) = 8}}} Now factor a 4 from the y-group.
{{{(3x^2)+4(y^2+2y) = 8}}} Now complete the square in the y-terms inside the parentheses by adding the square of half the y-coefficient or ({{{(2/2)^2 = 1}}}), remembering that you are really adding 4 because of the 4 in front of the parentheses.
{{{(3x^2)+4(y^2+2y+1) = 8+4}}} Now simplify this by factoring the trinomial in the y-terms.
{{{(3x^2)+4(y+1)^2 = 12}}} Now divide through by 12 to get the right side equal to 1.
{{{(3/12)x^2 + (4/12)(y+1)^2 = 12/12}}} Simplify this.
{{{x^2/4 + (y+1)^2/3 = 1}}}