Problem 17 - 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. $$ \frac{y^{2}}{36}-\frac{x^{2}}{4}=1 $$

Step-by-Step Solution

Answer

Vertices: (0, ±6); Foci: (0, ±2√10); Asymptotes: y = ±3x.

Step 1: Identify the type of hyperbola

The standard form of a hyperbola with a vertical opening is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). The given equation \( \frac{y^{2}}{36} - \frac{x^{2}}{4} = 1 \) matches this form, which confirms it opens vertically.

Step 2: Determine the values of 'a' and 'b'

From the equation \( \frac{y^{2}}{36} - \frac{x^{2}}{4} = 1 \), we have \( a^2 = 36 \) and \( b^2 = 4 \). Solving these gives us \( a = 6 \) and \( b = 2 \).

Step 3: Calculate the coordinates of the vertices

For a vertical hyperbola, the vertices are located at \( (0, \pm a) \). Thus, the vertices are \( (0, 6) \) and \( (0, -6) \).

Step 4: Calculate the coordinates of the foci

The formula to find the foci of a hyperbola is \( (0, \pm c) \), where \( c^2 = a^2 + b^2 \). Here, \( c^2 = 36 + 4 = 40 \) thus \( c = \sqrt{40} = 2\sqrt{10} \). Therefore, the foci are \( (0, 2\sqrt{10}) \) and \( (0, -2\sqrt{10}) \).

Step 5: Find the equations of the asymptotes

For a vertical hyperbola, the equations of the asymptotes are \( y = \pm \frac{a}{b}x \). Substituting \( a = 6 \) and \( b = 2 \), the equations become \( y = \pm 3x \).

Step 6: Graph the hyperbola

To graph the hyperbola, plot the vertices at \( (0, 6) \) and \( (0, -6) \). Draw the asymptote lines through the center at the origin with slopes \( 3 \) and \( -3 \). Sketch the hyperbola opening vertically, approaching the asymptotes.

Key Concepts

Vertices of Hyperbola

When we talk about the **vertices of a hyperbola**, we mean the points where the hyperbola intersects its transverse axis. These points are essential for understanding the shape and position of the hyperbola. For a vertical hyperbola, like the one given by the equation \( \frac{y^2}{36} - \frac{x^2}{4} = 1 \), the transverse axis is vertical. The formula to locate the vertices is \((0, \pm a)\).

Here, calculation led us to find \(a^2 = 36\), thus \(a = 6\). Therefore, the vertices are at coordinates \((0, 6)\) and \((0, -6)\).

These vertices help define the 'width' of the hyperbola along the transverse axis. To visualize, imagine them as the top and bottom tips of the vertical opening hyperbola.

Foci of Hyperbola

In a hyperbola, the **foci** are two points located inside the arms of the hyperbola. They are crucial for defining the hyperbola's shape since any point on the hyperbola has a constant difference in distance to these foci. For our equation, the foci lie on the transverse axis, just like the vertices.

The formula used is \((0, \pm c)\), where \(c\) is derived from \(c^2 = a^2 + b^2\). From our exercise, \(a^2 = 36\) and \(b^2 = 4\) yields \(c^2 = 40\), so \(c = \sqrt{40} = 2\sqrt{10}\).

Thus, the foci are positioned at \((0, 2\sqrt{10})\) and \((0, -2\sqrt{10})\). These points "pull" the arms of the hyperbola towards themselves, defining its unique curve shape.

Asymptotes of Hyperbola

The **asymptotes** of a hyperbola are lines that the hyperbola approaches but never touches. These provide a framework for the hyperbola's branches, offering insight into its overall behavior and direction. In a vertical hyperbola, the asymptotes intersect at the center of the hyperbola and run through the origin.

For our vertical hyperbola, the equations for the asymptotes are derived as \(y = \pm \frac{a}{b}x\). With \(a = 6\) and \(b = 2\), this results in \(y = \pm 3x\).

• The slopes of the asymptotes give us an idea about the steepness of the hyperbola's arms.
• These lines create a 'box' that guides how the hyperbola's branches curve upward and downward.

Understanding and visualizing these asymptote lines is key, as they chart the path that each branch of the hyperbola will follow, even though they will never intersect.