Problem 18 - 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. $$ \frac{y^{2}}{16}-\frac{x^{2}}{25}=1 $$

Step-by-Step Solution

Answer

Vertices: (0, ±4). Foci: (0, ±√41). Asymptotes: y = ±(4/5)x.

Step 1: Identify the Standard Form

The given equation is \( \frac{y^2}{16} - \frac{x^2}{25} = 1 \). This equation is already in the standard form of a hyperbola: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), indicating a vertical hyperbola with center at \( (0, 0) \). Here, \( a^2 = 16 \) and \( b^2 = 25 \).

Step 2: Calculate the Vertices

For a vertical hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the vertices are located at \( (0, \pm a) \). Calculating \( a \), we find \( a = \sqrt{16} = 4 \). Thus, the vertices are \( (0, 4) \) and \( (0, -4) \).

Step 3: Calculate the Foci

The foci of the hyperbola can be found using the formula \( c = \sqrt{a^2 + b^2} \). We know \( a^2 = 16 \) and \( b^2 = 25 \), so \( c = \sqrt{16 + 25} = \sqrt{41} \). The coordinates of the foci are \( (0, \pm \sqrt{41}) \).

Step 4: Find the Asymptotes Equations

The asymptotes for a vertical hyperbola are given by \( y = \pm \frac{a}{b}x \). Since \( a = 4 \) and \( b = 5 \), the equations of the asymptotes are \( y = \frac{4}{5}x \) and \( y = -\frac{4}{5}x \).

Step 5: Graph the Hyperbola

To graph the hyperbola, plot the center at \( (0, 0) \), the vertices at \( (0, 4) \) and \( (0, -4) \), and the foci at approximately \( (0, \pm 6.4) \) (using \( \sqrt{41} \approx 6.4 \)). Also, draw the asymptotes \( y = \frac{4}{5}x \) and \( y = -\frac{4}{5}x \). Sketch the hyperbola opening upwards and downwards along the \( y \)-axis, approaching the asymptotes but never crossing them.

Key Concepts

Vertices

The vertices of a hyperbola are pivotal points that provide insights into its structure. In the equation \( \frac{y^2}{16} - \frac{x^2}{25} = 1 \), we identify it as in the standard form of a vertical hyperbola. The vertices for a vertical hyperbola, described by the equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), are located at \( (0, \pm a) \). Here, \( a^2 = 16 \), so \( a = \sqrt{16} = 4 \).

• Vertices are the points where the hyperbola intersects its principal axis.
• For this equation, the vertices are \( (0, 4) \) and \( (0, -4) \).
• They define the width of the hyperbola along the vertical axis.

Understanding the position of vertices helps in sketching the hyperbola's overall framework, ensuring it opens and stretches correctly on the graph.

Foci

The foci of a hyperbola are two significant points that lie along its transverse axis. Unlike the vertices, foci are not directly visible on the hyperbola's curve, but they define its shape. To find the foci of the hyperbola \( \frac{y^2}{16} - \frac{x^2}{25} = 1 \), we use the formula \( c = \sqrt{a^2 + b^2} \). Here, \( a^2 = 16 \) and \( b^2 = 25 \), resulting in \( c = \sqrt{41} \) which is approximately 6.4.

• The foci lie along the "y" axis at \( (0, \pm \sqrt{41}) \).
• The hyperbola approaches the foci but never touches them, forming a curve around them.
• The distance between the center and each focus, \( c \), is crucial to the shape's elongation.

Foci are essential for hyperbole equations, as they influence the eccentricity and provide a complete understanding of the shape's geometry.

Asymptotes

Asymptotes are lines that a hyperbola approaches but never intersects, guiding its shape as it stretches to infinity. In the equation \( \frac{y^2}{16} - \frac{x^2}{25} = 1 \), the asymptotes are derived from the relationship between \( a \) and \( b \). As this is a vertical hyperbola, its asymptotes are represented by the equation \( y = \pm \frac{a}{b}x \). With \( a = 4 \) and \( b = 5 \), the asymptotes are \( y = \frac{4}{5}x \) and \( y = -\frac{4}{5}x \).

• Asymptotes intersect at the center of the hyperbola \( (0,0) \), setting the limit towards which the hyperbola extends.
• They create a cross, guiding the opening direction of the hyperbola.
• The slope of the asymptotes, \( \pm \frac{4}{5} \), indicates how steep the lines are, influencing the hyperbola's appearance.

Asymptotes play a crucial role in predicting the behavior of the hyperbola as it extends, helping to sketch and visualize its infinite paths properly.