Question

1. what are the foci of the ellipse given by the equation $x^2 - 4x + 5y^2 + 30y + 44 = 0$? (4, -3) and (0, -3) (2, -3) and (2, 3) (2, -3) and (-2, 3) (2, -1) and (2, -5)

Explanation:

Step1: Complete square for x-terms

$x^2 - 4x = (x-2)^2 - 4$

Step2: Complete square for y-terms

$5y^2 + 30y = 5(y^2 + 6y) = 5[(y+3)^2 - 9] = 5(y+3)^2 - 45$

Step3: Substitute into original equation

$(x-2)^2 - 4 + 5(y+3)^2 - 45 + 44 = 0$
Simplify: $(x-2)^2 + 5(y+3)^2 = 5$
Divide by 5: $\frac{(x-2)^2}{5} + \frac{(y+3)^2}{1} = 1$

Step4: Identify ellipse parameters

Standard form: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $(h,k)=(2,-3)$, $a^2=5$, $b^2=1$
Calculate $c$: $c = \sqrt{a^2 - b^2} = \sqrt{5-1} = 2$

Step5: Find foci coordinates

Foci lie along x-axis (since $a>b$): $(h\pm c, k)$

Answer:

(4, -3) and (0, -3)