Problem 15 - Precalculus Mathematics for Calculus

Question

Find the center, foci, vertices, and asymptotes of the hyperbola. Then sketch the graph. $$y^{2}-\frac{(x+1)^{2}}{4}=1$$

Step-by-Step Solution

Answer

Center: (-1, 0), Vertices: (-1, 1), (-1, -1), Foci: (-1, ±√5), Asymptotes: y = ±½(x + 1).

Step 1: Rearrange and Identify Hyperbola Type

First, rewrite the given equation of the hyperbola in the standard form. The provided equation is \( y^2 - \frac{(x+1)^2}{4} = 1 \). Re-arrange it to \[\frac{y^2}{1} - \frac{(x+1)^2}{4} = 1 \]This form corresponds to a hyperbola centered at \((h, k) = (-1, 0)\), where \(y\) is aligned with the transverse axis.

Step 2: Identify Key Elements of the Hyperbola

From the standard form, \(\frac{y^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), identify \(a^2=1\), \(b^2=4\). Therefore, \(a = 1\) and \(b = 2\). Set the center \( (h, k) = (-1, 0) \). Thus, the hyperbola has a vertical transverse axis.

Step 3: Find the Vertices

The vertices of the hyperbola are located a distance \(a\) from the center along the \(y\)-axis. Therefore, the vertices are at \[(h, k \pm a) = (-1, 0 \pm 1) = (-1, 1)\text{ and }(-1, -1).\]

Step 4: Find the Foci

For a hyperbola, the foci are located along the transverse axis at a distance \(c\), where \(c^2 = a^2 + b^2\). Thus:\[c^2 = 1 + 4 = 5 \quad \Rightarrow \quad c = \sqrt{5}\]The foci are located at:\[(h, k \pm c) = (-1, 0 \pm \sqrt{5}). \]

Step 5: Calculate the Asymptotes

The equations for the asymptotes for a hyperbola with a vertical transverse axis are given by:\[y - k = \pm \frac{a}{b}(x - h)\]Substituting the values:\[y - 0 = \pm \frac{1}{2}(x + 1)\text{or } y = \pm \frac{1}{2}(x + 1)\]This simplifies to: \[y = \frac{1}{2}(x + 1)\text{ and }y = -\frac{1}{2}(x + 1).\]

Step 6: Sketch the Hyperbola

To sketch the hyperbola, plot the center at \((-1, 0)\). Use the vertices \((-1,1)\) and \((-1, -1)\) and foci as guides. Draw the asymptotic lines through \((-1,0)\) with slopes \(\frac{1}{2}\) and \(-\frac{1}{2}\). Sketch the hyperbola opening upwards and downwards, asymptotically approaching these lines as \(x\) extends away from the center.

Key Concepts

center of hyperbola

The center of a hyperbola is a crucial point, as it acts as the reference from which other components like vertices and foci are determined. For the equation \( \frac{y^2}{1} - \frac{(x+1)^2}{4} = 1 \), the center is found from the standard hyperbola format \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). Here, \( (h, k) \) is the center. In our example, the terms simplify to indicate \( h = -1 \) and \( k = 0 \). This means the hyperbola's center is located at \((-1, 0)\).
The concept of the center is essential because it provides a foundational reference point for finding other symmetrical features of the hyperbola.

vertices of hyperbola

Vertices are important because they denote where the hyperbola intercepts its transverse axis and appears to be the 'tips' of the curve. For a vertical transverse axis, which we have here, the vertices are calculated by moving a distance \( a \) from the center along the \( y \)-axis. In our equation, \( a = 1 \).
Thus, starting from the center \((-1, 0)\), the vertices are found at \((-1, 1)\) and \((-1, -1)\). These points highlight how far the hyperbola stretches vertically from the center along its main axis of symmetry.

foci of hyperbola

The foci of a hyperbola are points located on the transverse axis, crucial for defining how 'spread out' the hyperbola appears. To find the foci, we use the formula \( c^2 = a^2 + b^2 \). In this example, we previously identified \( a^2 = 1 \) and \( b^2 = 4 \). Therefore, \( c^2 = 5 \), leading to \( c = \sqrt{5} \).
From the center \((-1, 0)\), the foci are calculated as \((-1, \sqrt{5})\) and \((-1, -\sqrt{5})\). These points signify significant locations within the structure of the hyperbola, influencing the shape and openness of the curves.

asymptotes of hyperbola

Asymptotes are lines that the hyperbola gets infinitely close to but never actually touches, forming a guideline for its 'arms' to approach. For hyperbolas with a vertical axis such as ours, the equations are \( y - k = \pm \frac{a}{b}(x - h) \). By substituting \( a = 1 \), \( b = 2 \), \( h = -1 \), and \( k = 0 \), the asymptote equations result in \( y = \frac{1}{2}(x + 1) \) and \( y = -\frac{1}{2}(x + 1) \).
These lines give a visual framework for sketching the hyperbola accurately, showing where it spreads but never crosses. Understanding asymptotes helps in capturing the hyperbola's infinite extension while maintaining its general shape.