Problem 49 - College Algebra

Question

For the following exercises, given information about the graph of the hyperbola, find its equation. Center: (4,2)\(;\) vertex: (9,2)\(;\) one focus: \((4+\sqrt{26}, 2)\)

Step-by-Step Solution

Answer

The equation is \(\frac{(x-4)^2}{25} - \frac{(y-2)^2}{1} = 1\).

Step 1: Understand the basics of a hyperbola

A hyperbola is defined as the set of all points in the plane where the difference in the distances to two fixed points (foci) is constant. A standard form for a horizontal hyperbola centered at \((h, k)\) is \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] where \((h, k)\) is the center and \(a\) is the distance from the center to a vertex.

Step 2: Determine the value of \(a\)

The vertex given is \((9,2)\) and the center is \((4,2)\). As the vertex is horizontal with respect to the center, the distance between the center and the vertex is \(a = 9 - 4 = 5\).

Step 3: Recognize the formula for the foci

The formula for the foci of a horizontal hyperbola is \((h\pm c, k)\). From the problem, one focus is given as \((4+\sqrt{26}, 2)\). This means \(c = \sqrt{26}\) and thus \((h+c, k)\) is a focus.

Step 4: Calculate \(b^2\) using \(c^2 = a^2 + b^2\)

We know \(c = \sqrt{26}\), so \(c^2 = 26\). We also know \(a = 5\), so \(a^2 = 25\). Using the formula \(c^2 = a^2 + b^2\), we get \(26 = 25 + b^2\). Solving for \(b^2\), we get \(b^2 = 26 - 25 = 1\).

Step 5: Write the equation of the hyperbola

Now that we have all necessary components, \(h = 4\), \(k = 2\), \(a = 5\), and \(b^2 = 1\), the equation of the hyperbola is \[\frac{(x-4)^2}{25} - \frac{(y-2)^2}{1} = 1\].

Key Concepts

Horizontal Hyperbola

A hyperbola consists of two separate curves called branches, and it can either open horizontally or vertically. In a horizontal hyperbola, the branches open to the left and right. The main defining feature is that the transverse axis (the line passing through the center and the vertices) is horizontal. Given our specific hyperbola in the problem, we know it's horizontal because the vertex and the center share the same y-coordinate, indicating movement along the x-axis. This is typical for a horizontal hyperbola where the vertices shift away from the center on a horizontal line. Knowing these characteristics helps in identifying the type of hyperbola and subsequently determining its equation.

Hyperbola Equation

The equation of a hyperbola depends on its orientation (horizontal or vertical). Here, we focus on the horizontal hyperbola. The formula for such a hyperbola centered at \( (h, k) \) is:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]

• \(h\) and \(k\) designate the x and y coordinates of the center.
• \(a\) is the distance from the center to a vertex horizontally.
• \(b^2\) is calculated based on \(c^2 = a^2 + b^2\), where \(c\) is the distance from the center to a focus.

This reflects the characteristic property of hyperbolas: the absolute difference in distances to the two foci (\(c\)) remains constant.When you have values for \(h, k, a^2\) and \(b^2\), you can directly plug them into this standard form equation of the horizontal hyperbola.

Foci of Hyperbola

Foci are two fixed points located symmetrically along the transverse axis of the hyperbola. For a horizontal hyperbola, the foci \( (h \pm c, k) \) extend beyond each vertex. In our example, one given focus is located at \( (4 + \sqrt{26}, 2) \), indicating that:

• \(h + c = 4 + \sqrt{26}\)
• Therefore, \(c = \sqrt{26}\)

These foci play a critical role in defining the hyperbola, as they determine distances used in the hyperbola's derivation. Understanding this, we can confirm how far the hyperbola stretches by determining the value of \(c\), linking it to the center of our curve for precise plotting.

Vertex of Hyperbola

The vertices of a hyperbola are critical points where the branches are closest or narrowest. For horizontal hyperbolas, vertices lie on the transverse axis. In this scenario, a vertex is given as \( (9,2) \), with the center purportedly at \( (4,2) \).

• The distance between the center and the vertex is \(a\).
• Calculate \(a\) by measuring the horizontal distance: here, it is \(9 - 4 = 5\).

This direct horizontal shift from the center to a vertex determines \(a\), further guiding the formation of the hyperbola's equation. Understanding each vertex's place helps describe the curve's structure and aids in the graph's accurate sketching.