Problem 38 - Beginning and Intermediate Algebra

Question

Identify the center of each hyperbola and graph the equation. $$4 y^{2}-x^{2}=16$$

Step-by-Step Solution

Answer

The center of the hyperbola \(4y^2 - x^2 = 16\) is at the origin (0, 0). To graph the hyperbola, first sketch its asymptotes \(y = \pm(1/2)x\), then plot the vertices at (0, -2) and (0, 2), and finally draw the branches of the hyperbola, making sure they are symmetric about the center and follow the asymptotes.

Step 1: Rewrite Equation in Standard Form

To rewrite the equation in standard form, we will first divide through by 16 on both sides: \[\frac{4y^2}{16} - \frac{x^2}{16} = 1\] Now, we can simplify the fractions: \[\frac{y^2}{4} - \frac{x^2}{16} = 1\] This is the standard form for a hyperbola (vertical orientation), where the equation is given by: \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]

Step 2: Identify the Center Coordinates

The coordinates of the center of the hyperbola are given by (h, k). Since there are no shifts in the given equation, the center is at the origin: Center: (0, 0)

Step 3: Find Values for a and b

In the standard equation, a^2 and b^2 represent horizontal and vertical scaling factors (in this case, we have vertical orientation of the hyperbola). In our simplified equation: a^2 = 4, so a = 2 b^2 = 16, so b = 4

Step 4: Find Vertices and Asymptotes of the Hyperbola

To find the vertices of the hyperbola, we will add and subtract the value "a" (which is 2) from the y-coordinate of the center (which is 0): Vertices: (0, ±2) To find the equations of the asymptotes of the hyperbola, we will use the standard formula for vertical hyperbolas: y = ±(a/b)x Using the values of "a" and "b" that we found before: y = ±(2/4)x y = ±(1/2)x

Step 5: Graph the Hyperbola

To graph the hyperbola, follow these steps: 1. Plot the center point (0, 0) on a coordinate plane. 2. Plot the vertices of the hyperbola at points (0, -2) and (0, 2). 3. Draw a rectangle centered at the hyperbola's center, using "2a" and "2b" as dimensions (4 units high by 8 units wide). 4. Draw the asymptotes by finding the slope of y = ±(1/2)x. 5. Sketch the hyperbola branches by plotting them symmetrically about the center, with their vertices on the y-axis, and following the asymptotes as guidelines. Following these steps will result in a graph of the hyperbola with the center at the point (0, 0), vertices at (0, ±2), and asymptotes y = ±(1/2)x.

Key Concepts

Standard form of a hyperbola

Understanding the standard form of a hyperbola is crucial to solving related equations and graphing them effectively. A hyperbola's standard form for a vertical orientation is expressed as:

• \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

This equation tells us several things about the hyperbola:

• The terms \((y-k)^2\) and \((x-h)^2\) indicate the squared differences from the center coordinates \((h, k)\).
• \(a^2\) and \(b^2\) are the denominators that help determine the shape and spread of the hyperbola.
• Notice the minus sign between the two fractions; this is a key feature of a hyperbola.

For a hyperbola with horizontal orientation, the form swaps the positions of \(x\) and \(y\), turning the equation into:

• \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)

Once the equation is in standard form, you can easily identify other aspects like the center, vertices, and asymptotes.

Center of a hyperbola

The center of a hyperbola is a significant coordinate point from which other properties of the hyperbola are derived. Identified as \((h, k)\), the center is located where both the \((y-k)^2\) and \((x-h)^2\) terms equate to zero. In their respective standard form equations, these variables indicate shifts in the hyperbola from the origin.

For our specific hyperbola equation,

• \( \frac{y^2}{4} - \frac{x^2}{16} = 1 \),
• there are no \(h\) and \(k\) values outside of the squares, so the hyperbola's center is directly at the origin, \((0, 0)\).

This indicates that the hyperbola sits symmetrically around this central point, aligning its vertices and following the asymptotes from here. Visualizing or plotting the center is usually the first step in graphing a hyperbola, ensuring that all other parts fall correctly into place around it.

Vertices of a hyperbola

Vertices are key points on a hyperbola that help outline its shape. In a vertically oriented hyperbola like ours, these points lie on the \(y\)-axis at a distance defined by \(a\) from the center.

Using our simplified equation:

• \(a^2 = 4\) implies \(a = 2\),

we determine that the vertices appear at \((0, \pm a)\) when the hyperbola is centered at the origin. For this particular curve:

• The vertices are \((0, 2)\) and \((0, -2)\).

These vertices essentially mark the closest and farthest extent of the branches of the hyperbola along the \(y\)-axis, giving a directional aim and structural limits.

Asymptotes of a hyperbola

Asymptotes in a hyperbola are slanted lines that the hyperbola approaches but never crosses. They serve as the boundaries within which the hyperbola's branches exist and guide their steepness and orientation.

• For vertically-oriented hyperbolas, the equation for asymptotes is given by \(y = \pm \frac{a}{b}x\).

Given our hyperbola's specific values of \(a = 2\) and \(b = 4\), the equation of the asymptotes becomes:

• \(y = \pm \frac{2}{4}x = \pm \frac{1}{2}x\).

These lines, passing through the center at \((0, 0)\), create a sort of invisible frame; the closer the hyperbola's branches get to them as they extend, the closer they "hug" these lines. Plotting the asymptotes on the graph is an essential step, serving as guides for accurately shaping the hyperbola's curves around the coordinate plane.