Problem 25 - Precalculus

Question

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

Step-by-Step Solution

Answer

Center: (1, -2), Vertices: (3, -2) and (-1, -2), Foci: (1 + \sqrt{5}, -2) and (1 - \sqrt{5}, -2), Asymptotes' equations: \(y = \frac{1}{2}x - \frac{5}{2}\) and \(y = -\frac{1}{2}x - \frac{1}{2}\)

Step 1: Find the center of the hyperbola

To find the center of the hyperbola, we simply look at the terms that are subtracted from x and y in the equation. Here, the center is \((h, k) = (1, -2)\)

Step 2: Find the vertices

The vertices of a hyperbola are \((h \pm a, k)\) for a horizontal hyperbola. Here, \(a^2 = 4\), so \(a = 2\). Therefore the vertices are \((1+2,-2) = (3,-2)\) and \((1-2,-2) = (-1,-2)\)

Step 3: Find the foci

The foci of a hyperbola are \((h \pm c, k)\) for a horizontal hyperbola. We find c using the equation \(c^2 = a^2 + b^2\). Here, \(b^2= 1\), so \(b=1\). Then \(c^2 = 4 + 1 = 5\) so \(c = \sqrt{5}\). The foci are \((1 + \sqrt{5}, -2)\) and \((1 - \sqrt{5}, -2)\)

Step 4: Find the equations of asymptotes

The equations of the asymptotes for a horizontal hyperbola are given by \(k = \pm \frac{b}{a}(x - h) + k\). Here, these become \(y = \pm \frac{1}{2}(x - 1) - 2\), which simplify to \(y = \frac{1}{2}x - \frac{5}{2}\) and \(y = -\frac{1}{2}x - \frac{1}{2}\)

Step 5: Sketch the hyperbola

We sketch the hyperbola by first drawing the center, vertices and foci. Then the asymptotes are drawn as diagonals of a rectangle formed by using a and b as half the lengths of the sides. The hyperbola is then drawn towards the vertices, approaching but never reaching the asymptotes

Key Concepts

Center of a Hyperbola

The center of a hyperbola is a critical point from which the other parts of the hyperbola are derived. It's quite similar to the center of a circle but serves a different purpose in hyperbolas. When you look at the standard form of a hyperbola equation, such as \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the center is identified by the \(h,k\) coordinates in the equation.
In the given hyperbola equation, \( \frac{(x-1)^{2}}{4}-\frac{(y+2)^{2}}{1}=1 \), the center point is found as \((1, -2)\) because \(x\) and \(y\) have been shifted by \(1\) and \(-2\) respectively. Knowing the center helps in sketching and understanding the positional aspects of a hyperbola.

Vertices of a Hyperbola

The vertices of a hyperbola are the points where each branch of the hyperbola intersects the transverse axis. For a standard horizontal hyperbola, the vertices are located at \( (h \pm a, k) \), where \(a\) is the square root of the term beneath the \(x\) part of the equation, in this case \(a^2 = 4\), so \(a=2\).
This means the vertices for our example are at \( (1 + 2, -2) = (3, -2) \) and \( (1 - 2, -2) = (-1, -2) \). Knowing the vertices can aid in sketching the hyperbola, as they are the points closest to and farthest from the center along the horizontal or vertical axis, depending on the orientation of the hyperbola.

Foci of a Hyperbola

The foci of a hyperbola are two points located along the transverse axis and are essential for defining the hyperbola's "shape." These points are located farther from the center than the vertices. The distance from the center to each focus is \(c\), where \(c^2 = a^2 + b^2\). In this example, we have \(a^2 = 4\) and \(b^2 = 1\), so \(c^2 = 5\), making \(c = \sqrt{5}\).
Thus, the foci are at \( (1 + \sqrt{5}, -2) \) and \( (1 - \sqrt{5}, -2) \). The foci contribute to the definition of the hyperbola, with the property that the absolute difference in distances from any point on the hyperbola to the two foci is constant.

Equations of Asymptotes

The asymptotes of a hyperbola are lines that provide a boundary the hyperbola will never touch or cross but will get infinitely close to. For a horizontal hyperbola, the equations of the asymptotes are \( y = k \pm \frac{b}{a}(x - h) \).
In our example, this gives us \( y = \pm \frac{1}{2}(x - 1) - 2 \) simplifying to \( y = \frac{1}{2}x - \frac{5}{2} \) and \( y = -\frac{1}{2}x - \frac{1}{2} \). Drawing these lines helps to sketch the hyperbola accurately, as the hyperbola branches will approach closer to these asymptotes without ever intersecting them. Understanding these equations is key for visualizing and drawing a hyperbola.