Problem 36 - College Algebra

Question

Sketch a graph of the hyperbola, labeling vertices and foci. \(\frac{(x-2)^{2}}{8}-\frac{(y+3)^{2}}{27}=1\)

Step-by-Step Solution

Answer

The hyperbola center is (2, -3) with vertices (2 ± 2√2, -3) and foci (2 ± √35, -3).

Step 1: Identify the Center

The given equation is \( \frac{(x - 2)^2}{8} - \frac{(y + 3)^2}{27} = 1 \). This indicates a hyperbola with the center at the point \( (h, k) = (2, -3) \).

Step 2: Determine Direction and Axes

The equation is in the form of \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), which tells us the hyperbola opens left and right (horizontally). Thus, the transverse axis (major axis) is horizontal.

Step 3: Calculate Vertices

For a hyperbola, the vertices are \( a \) units away from the center along the transverse axis. Here, \( a = \sqrt{8} = 2\sqrt{2} \). The vertices are at \( (2 \pm 2\sqrt{2}, -3) \).

Step 4: Calculate Foci

The distance to the foci from the center is given by \( c \), where \( c = \sqrt{a^2 + b^2} = \sqrt{8 + 27} = \sqrt{35} \). Thus, the foci are located at \( (2 \pm \sqrt{35}, -3) \).

Step 5: Sketch the Hyperbola

1. Plot the center at \( (2, -3) \).2. Plot the vertices at \( (2 + 2\sqrt{2}, -3) \) and \( (2 - 2\sqrt{2}, -3) \).3. Plot the foci at \( (2 + \sqrt{35}, -3) \) and \( (2 - \sqrt{35}, -3) \).4. Draw the transverse axis through the vertices.5. Sketch the hyperbola branches opening left and right from the vertices.

Key Concepts

Vertices

In a hyperbola, the vertices are key points that mark where each branch of the hyperbola is closest to the center. For the equation \(\frac{(x-2)^{2}}{8}-\frac{(y+3)^{2}}{27}=1\), the vertices lie along the transverse axis, which we've determined is horizontal. This means the vertices are located left and right of the center, which is \((2, -3)\).

To find the vertices' positions, identify the value \(a\) from the standard form of the hyperbola equation, where \(a^{2}\) is the denominator under the \((x-h)^{2}\) term. Here, \(a^{2} = 8\), so \(a = \sqrt{8} = 2\sqrt{2}\).

• Vertex 1: \((2 + 2\sqrt{2}, -3)\)
• Vertex 2: \((2 - 2\sqrt{2}, -3)\)

These vertices are symmetrically placed along the transverse axis, providing crucial reference points for sketching the hyperbola.

Foci

The foci of a hyperbola are points located along the transverse axis but beyond the vertices. The role of the foci is to determine the exact shape and "sharpness" of the hyperbola. For our equation \(\frac{(x-2)^{2}}{8}-\frac{(y+3)^{2}}{27}=1\), we calculate the foci using the formula for \(c\), the distance from the hyperbola's center to its foci: \[ c = \sqrt{a^{2} + b^{2}} \] where \(a^{2} = 8\) and \(b^{2} = 27\) give us:\[ c = \sqrt{8 + 27} = \sqrt{35} \]

• Focus 1: \((2 + \sqrt{35}, -3)\)
• Focus 2: \((2 - \sqrt{35}, -3)\)

These points are essential for understanding the "stretch" of the hyperbola's branches since the hyperbola is defined such that the difference in distances from any point on the hyperbola to the two foci is constant.

Transverse Axis

The transverse axis is a significant line because it contains both the center, the vertices, and runs through the foci of the hyperbola. For this specific equation \(\frac{(x-2)^{2}}{8}-\frac{(y+3)^{2}}{27}=1\), we recognize that the transverse axis is horizontal.

Here's how you can tell: the hyperbola's equation has the \(x\) part subtracted from the \(y\) part, i.e., it has the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). This specifies that the transverse axis aligns with the \(x\)-axis direction.

Key points about the transverse axis:

• It contains the center at \((2, -3)\).
• It runs through the vertex points \((2 \pm 2\sqrt{2}, -3)\).
• It is parallel to the x-axis, indicating the hyperbola opens left and right, not up and down.

Understanding the transverse axis helps with sketching the hyperbola correctly and positioning it on a coordinate plane.