Describe the hyperbola $y^2-\frac{\left(x-1\right)^2}{4}=1$.

• A. This is a vertical hyperbola centered at $\left(1,0\right)$ with vertices at $\left(1,1\right)$, $\left(1,-1\right)$ and foci at $\left(1,\sqrt5\right)$, $\left(1,-\sqrt5\right)$.
• B. This is a vertical hyperbola centered at $\left(1,0\right)$ with vertices at $\left(1,2\right),\left(1,-2\right)$ and foci at $\left(1,1\right),\left(1,-1\right)$.
• C. This is a horizontal hyperbola centered at $\left(-1,0\right)$ with vertices at $\left(0,0\right),\left(-2,0\right)$ and foci at $\left(\sqrt5-1,0\right),\left(-\sqrt5-1,0\right)$.
• D. This is a horizontal hyperbola centered at $\left(1,0\right)$ with vertices at $\left(0,0\right),\left(-2,0\right)$ and foci at $\left(1,-\sqrt5\right),\left(1,-\sqrt5\right)$.

Answer: (A)

Verified step by step guidance

1. Identify the standard form of a hyperbola equation. The given equation is \( y^2 - \frac{(x-1)^2}{4} = 1 \), which can be compared to the standard form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) for a vertical hyperbola.
2. Determine the center of the hyperbola. In the equation \( y^2 - \frac{(x-1)^2}{4} = 1 \), the center \((h, k)\) is \((1, 0)\) because the equation can be rewritten as \( \frac{(y-0)^2}{1} - \frac{(x-1)^2}{4} = 1 \).
3. Identify the vertices of the hyperbola. For a vertical hyperbola, the vertices are located at \((h, k \pm a)\). Here, \(a^2 = 1\), so \(a = 1\). Therefore, the vertices are \((1, 0 \pm 1)\), which are \((1, 1)\) and \((1, -1)\).
4. Calculate the foci of the hyperbola. The foci are located at \((h, k \pm c)\), where \(c^2 = a^2 + b^2\). Here, \(b^2 = 4\), so \(c^2 = 1 + 4 = 5\), giving \(c = \sqrt{5}\). Thus, the foci are \((1, 0 \pm \sqrt{5})\), which are \((1, \sqrt{5})\) and \((1, -\sqrt{5})\).
5. Summarize the characteristics of the hyperbola. This is a vertical hyperbola centered at \((1, 0)\) with vertices at \((1, 1)\) and \((1, -1)\), and foci at \((1, \sqrt{5})\) and \((1, -\sqrt{5})\).