Final answer:
The standard form of the ellipse with the given equation is 4(x - 2)² + 3(y - (3/2))² = -3/4.
Step-by-step explanation:
The standard form of an ellipse is given by the equation (x - h)²/a² + (y - k)²/b² = 1, where (h, k) is the center of the ellipse and a and b are the lengths of the semi-major and semi-minor axes respectively.
To put the equation 4x² + 3y² - 16x - 9y + 16 = 0 in standard form, we need to complete the square for both the x and y terms. First, let's rearrange the equation: 4x² - 16x + 3y² - 9y = -16.
Now, let's complete the square for the x terms: 4(x² - 4x) + 3y² - 9y = -16. We add and subtract the square of half the coefficient of x inside the parentheses: 4(x² - 4x + (4/2)²) + 3y² - 9y = -16 + 4(4/2)².
Simplifying, we get: 4(x - 2)² + 3y² - 9y = -16 + 4.
Now, let's complete the square for the y terms: 4(x - 2)² + 3(y² - 3y) = -12.
We add and subtract the square of half the coefficient of y inside the parentheses: 4(x - 2)² + 3(y² - 3y + (3/2)²) = -12 + 3(3/2)².
Simplifying, we get: 4(x - 2)² + 3(y - (3/2))² = -12 + 9/4.
The equation is now in standard form: 4(x - 2)² + 3(y - (3/2))² = -3/4.