The foci of the ellipse 9x2 + 4y2 –18x –24y + 9 = 0 are

To find the foci of the ellipse, we first need to rewrite the equation in standard form. The standard form of an ellipse is (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.

The given equation is 9x² + 4y² - 18x - 24y + 9 = 0.

First, group the x terms and the y terms:

9(x² - 2x) + 4(y² - 6y) = -9.

To complete the square, we take half of the coefficient of x and y, square it and add it to both sides. Half of -2 is -1, and (-1)² = 1. Half of -6 is -3, and (-3)² = 9.

So, we get:

9[(x - 1)² - 1] + 4[(y - 3)² - 9] = -9.

Simplify to get:

9(x - 1)² - 9 + 4(y - 3)² - 36 = -9.

Combine like terms:

9(x - 1)² + 4(y - 3)² = 36.

Divide through by 36 to get the equation in standard form:

(x - 1)²/4 + (y - 3)²/9 = 1.

So, the center of the ellipse (h,k) is (1,3), a = 2, and b = 3.

The distance from the center to each focus is given by the formula c = sqrt(a² - b²) = sqrt(4 - 9) = sqrt(-5).

Since the square root of a negative number is not a real number, there are no real foci for this ellipse.

To find the foci of the ellipse, we first need to rewrite the equation in standard form. The standard form of an ellipse is (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.

The given equation is 9x² + 4y² - 18x - 24y + 9 = 0.

First, group the x terms and y terms together:

9(x² - 2x) + 4(y² - 6y) = -9.

Next, complete the square for the x and y terms:

9[(x - 1)² - 1] + 4[(y - 3)² - 9] = -9.

Simplify this to get:

9(x - 1)² + 4(y - 3)² = 36.

Divide through by 36 to get the equation in standard form:

(x - 1)²/4 + (y - 3)²/9 = 1.

So, the center of the ellipse (h,k) is (1,3), a² = 4, and b² = 9.

The distance from the center to each focus is given by the formula c = sqrt(a² - b²) if a > b or c = sqrt(b² - a²) if b > a. Here, b > a, so we use the second formula:

c = sqrt(9 - 4) = sqrt(5).

So, the foci are at (h, k ± c) = (1, 3 ± sqrt(5)).