Question

find the center and foci of the ellipse: 9x^2 + 16y^2 + 126x + 96y + 441 = 0 center = (?, ) foci = ( ± √, )

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for x and y terms.

$$
LATEXBLOCK0
$$

Step2: Identify the center

For an ellipse in the form $\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1$, the center is $(h,k)$. Here $h=-7,k = - 3$, so the center is $(-7,-3)$.

Step3: Calculate $c$

We know that for an ellipse $c^{2}=a^{2}-b^{2}$, where $a^{2}=16$ and $b^{2}=9$. So $c=\sqrt{16 - 9}=\sqrt{7}$.

Step4: Find the foci

For an ellipse with a horizontal major - axis (since $a^{2}>b^{2}$, $a = 4,b = 3$), the foci are $(h\pm c,k)$. Substituting $h=-7,k=-3,c=\sqrt{7}$, the foci are $(-7\pm\sqrt{7},-3)$.

Answer:

Center = $(-7,-3)$
Foci = $(-7\pm\sqrt{7},-3)$