How to write x^2+(y^2)/4+(1/9)=1 in standard form or standard form of an ellipse?

Answer 1

To write x^2+(y^2)/4+(1/9)=1 in standard form of an ellipse, we need to simplify the equation by moving the constant term to the right side and dividing both sides by the constant term to make it 1:

x^2 + (y^2)/4 + (1/9) - 1 = 0

x^2 + (y^2)/4 - 8/9 = 0

To complete the square for y, we need to factor out the coefficient of y^2:

x^2 + (1/4)(y^2 - 8/2) = 0

x^2 + (1/4)(y^2 - 4) = 0

Now we can add and subtract (1/4) on the right side to complete the square for y:

x^2 + (1/4)(y^2 - 4) = 0

x^2 + (1/4)(y^2 - 4 + 4/4) = 0 + (1/4)(4/4)

x^2 + (1/4)(y - 0)^2 = 1/4

Finally, we can divide both sides by the constant term (1/4) to obtain the standard form of an ellipse:

(x - 0)^2 / (1/1)^2 + (y - 0)^2 / (2/1)^2 = 1

The center of the ellipse is (0,0), the lengths of the major and minor axes are 2 and 1, respectively.