Problem 37 - College Algebra

Question

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes \(\frac{(y+2)^{2}}{4}-\frac{(x-1)^{2}}{16}=1\)

Step-by-Step Solution

Answer

The center of the hyperbola is (1,-2), the vertices are located at (1, 0) and (1, -4), while the foci are located at points (1, -2+(2\sqrt{5})) and (1, -2-(2\sqrt{5})). The equations of the asymptotes will be \(y = -2 + \frac{(x - 1)}{2}\) and \(y = -2 - \frac{(x - 1)}{2}\).

Step 1: Identify Center, a, b

By comparing the given equation \( \frac{(y+2)^{2}}{4}-\frac{(x-1)^{2}}{16}=1\) with the general form of hyperbola equation \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1\) , the center of the hyperbola can be obtained by changing the signs of the coefficients of y and x inside the parenthesis. In this case, the center is at (1,-2). Also, a^2 = 4, hence a = 2; and b^2 = 16, hence b = 4.

Step 2: Find the Vertices

The vertices can be found from the center (h, k) by adding and subtracting the value of a to and from the y-coordinate of the center. Hence, the vertices are at (1, -2+2) and (1, -2-2), giving (1, 0) and (1, -4).

Step 3: Locate the Foci

The Foci can be found using the equation \(c = \sqrt{b^2 + a^2}\). Plugging in the values of a and b, we get \(c = \sqrt{16 + 4} = \sqrt{20}\), which simplifies to c = 2\sqrt{5}. The foci will then be at (1, -2+2\sqrt{5}) and (1, -2-2\sqrt{5}).

Step 4: Equation of the Asymptotes

The equations of asymptotes can be given as \(y = k \pm \frac{a}{b}(x-h)\). Substituting h = 1, k = -2, a = 2 and b = 4 in the equation, we get the equations of the asymptotes as \(y = -2 \pm \frac{1}{2}(x - 1)\). Hence, the equations of the asymptotes are \(y = -2 + \frac{1}{2}(x - 1)\) and \(y = -2 - \frac{1}{2}(x - 1)\).

Step 5: Graph the Hyperbola

Finally, plot all the identified points (center, vertices, foci), the hyperbola, and the asymptotes on the same graph. Start with the center of the hyperbola, then plot the vertices, and the foci. Sketch the asymptotes, then draw the hyperbola approaching these asymptotes.

Key Concepts

center of hyperbola

In the study of hyperbolas, the center is one of the first aspects you should identify. Imagine the center as the heart of the hyperbola. It's the point from which everything else is defined and coordinated.
To find the center of a hyperbola from its equation, compare it to the general form. For a hyperbola that opens vertically, the standard form is \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). Here, \((h, k)\) represents the center.

• In our example equation, \( \frac{(y+2)^2}{4} - \frac{(x-1)^2}{16} = 1 \), the center \((h, k)\) is located at \((1, -2)\).
• This is found by noting the values within the parentheses and changing their signs.

Understanding the center is crucial as it helps with plotting the hyperbola and finding its other features such as vertices and foci.

vertices of hyperbola

The vertices of a hyperbola are like the tips closest to the center. They define the shape and direction in which the hyperbola opens.
In a vertically oriented hyperbola (as in our example), the vertices lie directly above and below the center along the y-axis.

• To find the vertices, you add and subtract the value of \(a\) to and from the y-coordinate of the center.
• Given our hyperbola's center at \((1, -2)\) and \(a = 2\) (since \(a^2 = 4\)), the vertices will be located at \((1, 0)\) and \((1, -4)\).

It is by identifying the vertices that you first start shaping the actual graph of the hyperbola on paper, providing you with a visual guide for plotting the structure.

foci of hyperbola

The foci of a hyperbola are special points that help define the shape and the elongation of the hyperbola. Like the vertices, they lie along the axis of symmetry but are always farther from the center than the vertices.
The formula for finding the foci of a hyperbola is different from that of ellipses. For a vertically oriented hyperbola like ours, calculate \(c\) using the equation \(c = \sqrt{b^2 + a^2}\).

• In our equation, \(b^2 = 16\) and \(a^2 = 4\), giving \(c = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\).
• The foci are then at \((1, -2 + 2\sqrt{5})\) and \((1, -2 - 2\sqrt{5})\).

These foci help define that distinctive open ">" or "<" shape of the hyperbola, with each curve bowing outward around each focus.

asymptotes of hyperbola

Asymptotes in a hyperbola act as invisible boundaries that the arms of the hyperbola approach but never touch. They help us understand how the curve will behave as it extends toward infinity.
To find the equations of the asymptotes for our vertical hyperbola, use the equation \(y = k \pm \frac{a}{b}(x-h)\), where \((h, k)\) is the center, and \(a\) and \(b\) are the values determined from the equation.

• In this case, substitute \(h = 1\), \(k = -2\), \(a = 2\), and \(b = 4\) into the formula.
• This yields the equations \(y = -2 \pm \frac{1}{2}(x - 1)\).
• Explicitly, these are \(y = -2 + \frac{1}{2}(x - 1)\) and \(y = -2 - \frac{1}{2}(x - 1)\).

The asymptotes give the hyperbola its guiding contour, displaying how the arms of the hyperbola open outward, frameworking the entire graph. Understanding asymptotes is crucial for accurately sketching the hyperbola.