Problem 15 - Precalculus with Limits

Question

In Exercises 9-22, find the center, vertices, foci, and the equations of the asymptotes of the hyperbola, and sketch its graph using the asymptotes as an aid. \(\dfrac{(x-1)^2}{4}-\dfrac{(y+2)^2}{1}=1\)

Step-by-Step Solution

Answer

The center is at (1, -2). Vertices are at (3,-2) and (-1,-2). Foci are located at (1+\(\sqrt{5}\), -2) and (1-\(\sqrt{5}\), -2). Equations of the asymptotes are \(y+2= \pm \frac{1}{2}(x-1)\).

Step 1: Identify center, a and b

From the given hyperbola equation \(\dfrac{(x-1)^2}{4}-\dfrac{(y+2)^2}{1}=1\), we can clearly identify the center (h, k) as (1, -2). 'a' and 'b' (constants representing half the lengths of the Major and Minor axis respectively) can also be identified. 'a' is \(\sqrt{4}\)=2 and 'b' is \(\sqrt{1}\)=1.

Step 2: Find c (distance from center to foci)

From the identified values of 'a' and 'b', find 'c' using the equation \(c=\sqrt{a^2+b^2}\). Hence, \(c=\sqrt{2^2+1^2}=\sqrt{5}\). This marks the distance of foci from the center.

Step 3: Find the vertices

To find the vertices, add and subtract 'a' from the x-coordinate of the center as the horizontal axis is the major axis here. So, the vertices are (1+2,-2) and (1-2, -2), i.e., (3,-2) and (-1,-2).

Step 4: Find the foci

Subtract and add 'c' (found in Step 2) from x-coordinate of the center to get the foci. The foci then are at positions (1+\(\sqrt{5}\), -2) and (1-\(\sqrt{5}\), -2).

Step 5: Find equations of the asymptotes

Use the equation for the asymptotes \(y = k \pm \frac{b}{a}(x - h)\). Plug in the values for 'h', 'k', 'a', and 'b'. We get the equations of the asymptotes as \(y+2= \pm \frac{1}{2}(x-1)\).

Step 6: Sketch the graph

Plot the center, vertices, and foci on the graph. The hyperbola opens to left and right as the x term is positive in the equation. Sketch the asymptotes and draw the opening of hyperbola along these lines.

Key Concepts

Equation of Asymptotes

In any hyperbola, the asymptotes are crucial guidelines that help us sketch the graph and understand its limits. For the equation \( \frac{(x-1)^2}{4} - \frac{(y+2)^2}{1} = 1 \), we know that the hyperbola is centered at \((h, k) = (1, -2)\) and its transverse axis is along the x-axis, making it a horizontal hyperbola. The general formula for the asymptotes of a horizontal hyperbola is:

• \( y = k \pm \frac{b}{a}(x - h) \)

Here, \('b'\) is the half-length of the semi-minor axis, and \('a'\) is the half-length of the semi-major axis.
For our specific hyperbola, \(a = \sqrt{4} = 2\) and \(b = \sqrt{1} = 1\). Substituting these values into the formula gives:

• \(y + 2 = \pm \frac{1}{2}(x - 1)\)

This equation tells us that the hyperbola has slanted asymptotic lines that help it widen around its center point. These imaginary lines do not intersect the curve but frame the path the hyperbola approaches as it spreads out.

Vertices of a Hyperbola

The vertices of a hyperbola are pivotal points that denote the furthest reaches of the hyperbola along its transverse axis. For the given equation \( \frac{(x-1)^2}{4} - \frac{(y+2)^2}{1} = 1 \), we identify the center of the hyperbola as \((1, -2)\). The positive value under the x-term indicates a horizontal opening.
To find the vertices, you add and subtract \('a'\), the distance from the center to a vertex, from the x-coordinate of the center, because the x-axis is the transverse axis for this hyperbola. With \(a = 2\), the vertices are calculated as follows:

• \((1 + 2, -2) = (3, -2)\)
• \((1 - 2, -2) = (-1, -2)\)

The vertices \((3, -2)\) and \((-1, -2)\) highlight the width of the hyperbola's opening along the x-axis. This helps to accurately map the curve's path in a graph.

Foci of a Hyperbola

The foci of a hyperbola are special points that lie along the transverse axis, further out than the vertices, and determine the shape and direction of the hyperbola. For the given hyperbola equation \( \frac{(x-1)^2}{4} - \frac{(y+2)^2}{1} = 1 \), we first find the value of \('c'\), the distance from the center to each focus. The formula to calculate \(c\) is:

• \(c = \sqrt{a^2 + b^2}\)

Given \(a = 2\) and \(b = 1\), substitute these to get:

• \(c = \sqrt{2^2 + 1^2} = \sqrt{5}\)

With the center at \((1, -2)\) and using \(c\), the foci positions are:

• \((1 + \sqrt{5}, -2)\)
• \((1 - \sqrt{5}, -2)\)

These calculations reveal the points \((1 + \sqrt{5}, -2)\) and \((1 - \sqrt{5}, -2)\), where the actual foci reside. The foci act as guiding points for the hyperbola's curving arms.

Center of a Hyperbola

The center of a hyperbola is a fundamental point from which all other characteristics, like vertices and foci, are determined. In any equation of the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the center of the hyperbola is given as \((h, k)\).
For the example equation \( \frac{(x-1)^2}{4} - \frac{(y+2)^2}{1} = 1 \), we can directly read the center \((h, k)\) as \((1, -2)\). This central point is not only the midpoint of the vertices and the foci but also the foundational axis around which the graph pivots.
Knowing the center helps in efficiently finding other elements of the hyperbola, such as distances to vertices and foci, and crafting the graph's precise symmetry.