Question

find the center, foci, and vertices of the ellipse.
\\(\frac{(x + 1)^2}{36} + \frac{(y - 1)^2}{25} = 1\\)
select a single answer

• center at \\((-1,1)\\) foci at \\((-1 + sqrt{11}, -1), (-1 - sqrt{11}, -1)\\) vertices at \\((6,1), (-6,1)\\)
• center at \\((-1,1)\\) foci at \\((-sqrt{11},1), (sqrt{11},1)\\) vertices at \\((6,1), (-6,1)\\)
• center at \\((-1,1)\\) foci at \\((-1 + sqrt{11},1), (-1 - sqrt{11},1)\\) vertices at \\((-7,1), (5,1)\\)

Explanation:

Step1: Recall ellipse standard form

The horizontal major axis ellipse form is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ where $a>b$, center $(h,k)$.

Step2: Identify center $(h,k)$

Compare $\frac{(x+1)^2}{36}+\frac{(y-1)^2}{25}=1$ to standard form: $h=-1$, $k=1$.
Center: $(-1,1)$

Step3: Calculate $a, b, c$

$a^2=36 \implies a=6$; $b^2=25 \implies b=5$.
Use $c^2=a^2-b^2$:
$c^2=36-25=11 \implies c=\sqrt{11}$

Step4: Find vertices

Vertices: $(h\pm a,k) = (-1\pm6,1)$
$(-1+6,1)=(5,1)$; $(-1-6,1)=(-7,1)$

Step5: Find foci

Foci: $(h\pm c,k) = (-1\pm\sqrt{11},1)$
$(-1+\sqrt{11},1)$; $(-1-\sqrt{11},1)$

Answer:

center at $(-1,1)$ foci at $(-1+\sqrt{11},1), (-1-\sqrt{11},1)$ vertices at $(-7,1),(5,1)$