Problem 21 - Algebra 2

Question

Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation. Then graph the hyperbola. $$ \frac{(y-3)^{2}}{25}-\frac{(x-2)^{2}}{16}=1 $$

Step-by-Step Solution

Answer

Vertices: (2, 8) and (2, -2); Foci: (2, 3±√41); Asymptotes: y = 3±(5/4)(x-2).

Step 1: Identify the Standard Form

The given equation of a hyperbola is \( \frac{(y-3)^2}{25} - \frac{(x-2)^2}{16} = 1 \). This equation resembles the standard form of a vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). From this, we can see that \( (h, k) = (2, 3) \), \( a^2 = 25 \), and \( b^2 = 16 \).

Step 2: Calculate the Vertices

For a vertical hyperbola, the vertices are found at \( (h, k \pm a) \). Since \( a = \sqrt{25} = 5 \), the vertices are \( (2, 3 \pm 5) = (2, 8) \) and \( (2, -2) \).

Step 3: Find the Foci

The foci of the hyperbola are determined using the formula \( c^2 = a^2 + b^2 \). Here, \( c^2 = 25 + 16 = 41 \), hence \( c = \sqrt{41} \). So, the foci are at \( (h, k \pm c) = (2, 3 \pm \sqrt{41}) \).

Step 4: Determine the Asymptotes

For this vertical hyperbola, the asymptotes are given by the equations \( y = k \pm \frac{a}{b}(x-h) \). Substituting the values, we get \( y = 3 \pm \frac{5}{4}(x-2) \). This gives us two asymptote equations: \( y = 3 + \frac{5}{4}(x-2) \) and \( y = 3 - \frac{5}{4}(x-2) \).

Step 5: Graph the Hyperbola

To graph the hyperbola, plot the center at \((2, 3)\), the vertices at \((2, 8)\) and \((2, -2)\), and the foci at approximately \((2, 3 + 6.4)\) and \((2, 3 - 6.4)\). Draw the asymptotes as diagonal lines intersecting at the center. Sketch the hyperbola with branches opening vertically, approaching but never touching the asymptotes.

Key Concepts

Vertices of Hyperbolas

In the context of hyperbolas, vertices represent critical points that lie on the main axis of the hyperbola. These points are crucial as they help define the hyperbola's shape and direction.
For a vertical hyperbola like the given one, which has the equation \( \frac{(y-3)^2}{25} - \frac{(x-2)^2}{16} = 1 \), the vertices are determined using the center \((h, k)\) and the value of \(a\).
Here, the center is at \((2, 3)\) and \(a = \sqrt{25} = 5\).

• The vertices are located at \((h, k \pm a)\).
• This results in the coordinates \((2, 8)\) and \((2, -2)\).

The vertices indicate the points where the hyperbola intersects the transverse axis, the axis along which the hyperbola opens.

Foci of Hyperbolas

The foci of a hyperbola are points that reside within each branch of the hyperbola, just beyond the vertices. They play a foundational role in defining the hyperbola's curves.
To find the foci for the hyperbola given by \( \frac{(y-3)^2}{25} - \frac{(x-2)^2}{16} = 1 \), we use the relationship \( c^2 = a^2 + b^2 \).

• Here, \(a^2 = 25\) and \(b^2 = 16\).
• Therefore, \(c^2 = 25 + 16 = 41\), leading to \(c = \sqrt{41}\).

Subsequently, the foci are placed at \((h, k \pm c) = (2, 3 \pm \sqrt{41})\).
The foci are significant because the difference in distances from any point on the hyperbola to the foci is constant.

Equations of Asymptotes

Asymptotes are oblique lines that the hyperbola approaches but never intersects or crosses. These lines serve as guides to help shape the graph of a hyperbola.
For the hyperbola in question, characterized by the equation \( \frac{(y-3)^2}{25} - \frac{(x-2)^2}{16} = 1 \), the asymptotes have equations based on the slopes derived from \( \frac{a}{b} \).

• Since \(a = 5\) and \(b = 4\), the slopes of the asymptotes are \(\pm \frac{5}{4}\).
• The equations of the asymptotes are: \[ y = 3 \pm \frac{5}{4}(x-2) \].

These equations help visualize the steepness and direction of the asymptotes, which dictate how the hyperbola's branches curve as they extend towards infinity.

Graphing Hyperbolas

Graphing a hyperbola involves several steps, weaving together all the elements like the center, vertices, foci, and asymptotes. To effectively graph the given hyperbola:

• Start by plotting the center at \((2, 3)\).
• Next, add the vertices at \((2, 8)\) and \((2, -2)\).
• Position the foci approximately at \((2, 3 + 6.4)\) and \((2, 3 - 6.4)\) using the previously calculated \(c = \sqrt{41} \approx 6.4\).
• Draw the asymptotic lines based on the equations \(y = 3 \pm \frac{5}{4}(x-2)\), which will pass through the center.

Once these elements are on the graph, sketch the hyperbola's branches. They should open vertically and curve towards the asymptotes but never touch them.
The sketch reveals the distinctive twin curve shape of the hyperbola, illustrating its mathematical properties visually.