Problem 15 - Algebra and Trigonometry

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$ 25 y^{2}-9 x^{2}=225 $$

Step-by-Step Solution

Answer

Vertices: (0, 3), (0, -3); Foci: (0, √34), (0, -√34); Asymptotes: y = ±(3/5)x.

Step 1: Standard Form

The given hyperbola equation is \( 25y^2 - 9x^2 = 225 \). First, we divide the entire equation by 225 to transform it into its standard form: \[ \frac{y^2}{9} - \frac{x^2}{25} = 1 \]. This reveals it's a vertical hyperbola (because the \( y^2 \) term comes first).

Step 2: Identify the Vertices

For the standard form equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the vertices of a vertical hyperbola are located at \((0, \pm a)\). In our equation \( a^2 = 9 \), so \( a = 3 \). Therefore, the vertices are \( (0, 3) \) and \( (0, -3) \).

Step 3: Calculate the Foci

The distance to the foci \( c \) can be calculated using the formula \( c = \sqrt{a^2 + b^2} \). Here, \( b^2 = 25 \), so \( c = \sqrt{9 + 25} = \sqrt{34} \). Therefore, the foci are at \( (0, \sqrt{34}) \) and \( (0, -\sqrt{34}) \).

Step 4: Determine the Asymptotes

The equations of the asymptotes for a vertical hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) are given by \( y = \pm \frac{a}{b}x \). Substituting \( a = 3 \) and \( b = 5 \), the asymptotes are \( y = \pm \frac{3}{5}x \).

Step 5: Sketch the Graph

For sketching, plot the center at \((0,0)\), vertices at \((0,3)\) and \((0,-3)\), foci at \((0,\sqrt{34})\) and \((0,-\sqrt{34})\), and draw the asymptotes \( y = \frac{3}{5}x \) and \( y = -\frac{3}{5}x \). The hyperbola will open vertically around these guidelines.

Key Concepts

Vertices of a Hyperbola

In the context of hyperbolas, vertices are crucial as they represent the points where each branch of the hyperbola is closest to its center. To find the vertices of a hyperbola in the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), we utilize the variable \(a\). For a vertical hyperbola, the vertices are located at \((0, \pm a)\). This makes understanding the orientation essential, as it indicates how the hyperbola opens.
In the given exercise, we identified that \( a^2 = 9 \), so \( a = 3 \). Therefore, the vertices are located at the points \((0, 3)\) and \((0, -3)\). These points should always align vertically through the center of the hyperbola when plotted on a graph.

Foci of a Hyperbola

The foci of a hyperbola are two points, each located at a specific distance from the center, which help define the hyperbola's shape. For a vertical hyperbola like ours, the foci are positioned along the y-axis at \((0, \pm c)\).
To find \(c\), we use the formula \(c = \sqrt{a^2 + b^2}\). In solving our problem, we found \(b^2 = 25\) and consequently \(c = \sqrt{34}\). So, the foci are \((0, \sqrt{34})\) and \((0, -\sqrt{34})\). These points are further apart from the center than the vertices, and they give the hyperbola its characteristic shape.

Asymptotes of a Hyperbola

Asymptotes in a hyperbola are lines that the hyperbola approaches but never actually reaches. For the equation \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the asymptotes help define the direction in which the branches of the hyperbola open as they extend to infinity.
For a vertical hyperbola, the equations of the asymptotes are given by \(y = \pm \frac{a}{b}x\). Substituting the values from our exercise, where \(a = 3\) and \(b = 5\), we find the asymptotes are \(y = \frac{3}{5}x\) and \(y = -\frac{3}{5}x\). These lines create a cross through the center and guide the shape of the hyperbola.

Graphing Hyperbolas

Graphing a hyperbola involves carefully placing its center, vertices, foci, and asymptotes on the coordinate plane. For the exercise we have, the center is at \((0,0)\). From there, plot the vertices at \((0,3)\) and \((0,-3)\), as these are the closest points the hyperbola gets to the center.
Next, mark the foci at \((0, \sqrt{34})\) and \((0, -\sqrt{34})\). These help in visualizing the extent and orientation of the hyperbola's branches. Draw the asymptotes \(y = \frac{3}{5}x\) and \(y = -\frac{3}{5}x\); these lines act like invisible boundaries directing how the hyperbola's limbs stretch outward.
Finally, sketch the hyperbola's two branches approaching but never touching the asymptotes, reflecting through the plotted points and consistently following the curves defined by the foci and vertices.