Determine the vertices and foci of the ellipse $\left(x+1\right)^2+\frac{\left(y-2\right)^2}{4}=1$.

• A. Vertices: $\left(-1,4\right),\left(-1,0\right)$Foci: $\left(-1,2+\sqrt3\right),\left(-1,2-\sqrt3\right)$
• B. Vertices: $\left(-1,4\right),\left(-1,0\right)$Foci: $\left(-2,2\right),\left(0,2\right)$
• C. Vertices: $\left(-2,2\right),\left(0,2\right)$Foci: $\left(1,2+\sqrt3\right),\left(1,2-\sqrt3\right)$
• D. Vertices: $\left(-2,2\right),\left(0,2\right)$Foci: $\left(2+\sqrt3,1\right),\left(2-\sqrt3,1\right)$

Answer: (A)

Verified step by step guidance

1. Identify the standard form of the ellipse equation. The given equation is \((x+1)^2 + \frac{(y-2)^2}{4} = 1\). This is in the form \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\), where \(a > b\).
2. Determine the center of the ellipse. From the equation, \(h = -1\) and \(k = 2\), so the center is \((-1, 2)\).
3. Identify the values of \(a\) and \(b\). Here, \(a^2 = 4\) and \(b^2 = 1\), so \(a = 2\) and \(b = 1\). Since \(a > b\), the major axis is vertical.
4. Calculate the vertices. For a vertical ellipse, the vertices are at \((h, k \pm a)\). Thus, the vertices are \((-1, 2 + 2)\) and \((-1, 2 - 2)\), which simplifies to \((-1, 4)\) and \((-1, 0)\).
5. Find the foci. The distance from the center to each focus is \(c\), where \(c^2 = a^2 - b^2\). Calculate \(c = \sqrt{4 - 1} = \sqrt{3}\). The foci are at \((h, k \pm c)\), which are \((-1, 2 + \sqrt{3})\) and \((-1, 2 - \sqrt{3})\).