Problem 16 - Precalculus Mathematics for Calculus

Question

Find the center, foci, vertices, and asymptotes of the hyperbola. Then sketch the graph. $$\frac{(y-1)^{2}}{25}-(x+3)^{2}=1$$

Step-by-Step Solution

Answer

Center: \((-3,1)\), Vertices: \((-3,6)\), \((-3,-4)\), Foci: \((-3,1\pm\sqrt{26})\), Asymptotes: \(y=5x+16\), \(y=-5x-14\).

Step 1: Identify the Form of the Hyperbola

The given equation is \(\frac{(y-1)^2}{25} - (x+3)^2 = 1\). This has the form \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), which represents a vertical hyperbola. Here, \(a^2 = 25\) and \(b^2 = 1\).

Step 2: Determine the Center

For the hyperbola, the center is at the point \((h, k)\). From the equation \(\frac{(y-1)^2}{25} - \frac{(x+3)^2}{1} = 1\), we have \(h = -3\) and \(k = 1\). So, the center is at \((-3, 1)\).

Step 3: Find the Vertices

For a vertical hyperbola, the vertices are \((h, k \pm a)\). Here, \(a = \sqrt{25} = 5\). Thus, the vertices are at \((-3, 1 \pm 5)\) or \((-3, 6)\) and \((-3, -4)\).

Step 4: Find the Foci

The foci of a vertical hyperbola are at \((h, k \pm c)\), where \(c = \sqrt{a^2 + b^2}\). Hence, \(c = \sqrt{25 + 1} = \sqrt{26}\). So, the foci are at \((-3, 1 \pm \sqrt{26})\).

Step 5: Determine the Asymptotes

For a vertical hyperbola, the asymptotes are given by the equations \(y = k \pm \frac{a}{b}(x - h)\). Here, \(\frac{a}{b} = \frac{5}{1} = 5\). So, the equations of the asymptotes are \(y = 1 + 5(x + 3)\) and \(y = 1 - 5(x + 3)\), which simplify to \(y = 5x + 16\) and \(y = -5x - 14\).

Step 6: Sketch the Graph

Plot the center \((-3, 1)\), vertices \((-3, 6)\) and \((-3, -4)\), and draw the asymptotes \(y = 5x + 16\) and \(y = -5x - 14\). Sketch the branches of the hyperbola opening vertically, approaching but never touching the asymptotes.

Key Concepts

Conic Sections

Conic sections are curves obtained by intersecting a plane with a double napped cone. They are fundamental concepts in mathematics and their study is essential in understanding more advanced geometric properties. The main types of conic sections are:

• Circle - A set of points equidistant from a central point.
• Ellipse - Similar to a circle but stretched along two axes.
• Parabola - A symmetric curve with a single peak or trough.
• Hyperbola - A curve with two separate branches.

Hyperbolas occur when the cutting plane intersects both halves of the cone but is not parallel to the cone's base. Each form of a hyperbola has distinct definitions and properties which help in identifying and graphing them.

Vertices

Vertices are key points on a hyperbola. They represent the points where each branch of the hyperbola is closest to the center. For hyperbolas, there are always two vertices, one on each branch.

• In a vertical hyperbola, these vertices lie above and below the center.
• In a horizontal hyperbola, they lie left and right of the center.

The location of the vertices is determined by the variable 'a' in the hyperbola's standard form equation. For a vertical hyperbola like \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\):
- The vertices are located at \((h, k \pm a)\).
Understanding how to find these points is vital as they help in sketching the curve and serve as a guide for locating other key features like the foci and asymptotes.

Foci

The foci (plural of focus) are another set of points that define a hyperbola's properties. These points are located inside each branch of the hyperbola. They play a crucial role in understanding the shape and behavior of the hyperbola.

• For a vertical hyperbola, the foci are positioned vertically relative to the center.
• For a horizontal hyperbola, they are positioned horizontally.

Foci can be calculated using the formula \(c = \sqrt{a^2 + b^2}\), where \(c\) is the distance from the center to each focus. In our given equation, the foci would be at \((h, k \pm c)\), where \(c\) is determined by the lengths of \(a\) and \(b\). Knowing the location of the foci helps in understanding the direction and spread of the hyperbola's branches.

Asymptotes

Asymptotes are lines that a curve approaches as it heads towards infinity.In hyperbolas, these lines provide crucial information about the orientation and the "opening style" of the hyperbola branches.

• These lines are straight and do not intercept the hyperbola.
• They usually cross at the hyperbola's center.

In a vertical hyperbola with the form \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), the equations for the asymptotes are:
\(y = k \pm \frac{a}{b}(x-h)\).
This is key for sketching the hyperbola accurately, ensuring you understand how and where the branches approach these lines, especially in a graph.Finding the asymptotes helps create a frame that ensures you draw the hyperbola's shape accurately, reflecting its true geometric properties.