Problem 15 - Precalculus Mathematics for Calculus

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 = ±3/5x.

Step 1: Identify the Standard Form of the Hyperbola

The given equation is \(25y^2 - 9x^2 = 225\). First, we need to rewrite it in the standard hyperbola equation form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Divide the entire equation by 225 to achieve this: \(\frac{25y^2}{225} - \frac{9x^2}{225} = \frac{225}{225}\). Simplify to get \(\frac{y^2}{9} - \frac{x^2}{25} = 1\). This equation represents a hyperbola centered at the origin with \(a^2 = 9\) and \(b^2 = 25\).

Step 2: Find the Vertices of the Hyperbola

For a hyperbola in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices are located at \( (0, \pm a) \). Given \(a^2 = 9\), find \(a\) by calculating \(a = \sqrt{9} = 3\). Thus, the vertices are \((0, 3)\) and \((0, -3)\).

Step 3: Calculate the Foci of the Hyperbola

The formula for the foci of a hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) is \((0, \pm c)\) where \(c = \sqrt{a^2 + b^2}\). Use the given values \(a^2 = 9\) and \(b^2 = 25\) to find \(c\): \(c = \sqrt{9 + 25} = \sqrt{34}\). Therefore, the foci are \((0, \sqrt{34})\) and \((0, -\sqrt{34})\).

Step 4: Determine the Asymptotes of the Hyperbola

For the hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the equations of the asymptotes are \(y = \pm \frac{a}{b}x\). Here, \(a = 3\) and \(b = 5\). Therefore, the asymptotes are \(y = \pm \frac{3}{5}x\).

Step 5: Sketch the Graph of the Hyperbola

With the vertices \((0, 3)\) and \((0, -3)\), foci \((0, \sqrt{34})\) and \((0, -\sqrt{34})\), and asymptotes \(y = \pm \frac{3}{5}x\), sketch the hyperbola. Draw the center at the origin, plot the vertices along the y-axis, and use the slopes of the asymptotes to sketch the extended lines crossing through the origin. The hyperbola will open upwards and downwards along the y-axis, curving towards its asymptotes.

Key Concepts

Vertices of a Hyperbola

The vertices of a hyperbola are fundamental points from which the curve seems to 'begin' and 'end'. These points are crucial because they help define the shape and orientation of the hyperbola.
To determine the vertices, first ensure the equation of the hyperbola is in the standard form. For our example, the equation is \[\frac{y^2}{9} - \frac{x^2}{25} = 1\].

• Here, the vertices can be found along the axis which corresponds to the positive term in the equation. In our case, this is the y-axis, since \(y^2\) is divided by \(a^2 = 9\).
• Calculated, \(a = \sqrt{9} = 3\), hence the vertices are at \((0, \pm 3)\). This means the hyperbola will extend vertically from \(-3\) to \(3\) on the y-axis.

These points serve as a guide to sketching the hyperbola, emphasizing its vertical stretch.

Foci of a Hyperbola

The foci of a hyperbola are special points that are always located within the curve. They are essential for understanding the hyperbola because the difference in distances from any point on the hyperbola to the foci is constant. This intrinsic property makes hyperbolas unique.
To find the foci, we use the equation \(c = \sqrt{a^2 + b^2}\).

• Given \(a^2 = 9\) and \(b^2 = 25\), we find \(c = \sqrt{9 + 25} = \sqrt{34}\).
• The foci are positioned along the same axis as the vertices, given the equation's form. Thus, they are located at \((0, \pm \sqrt{34})\).

These foci lie further along the y-axis than the vertices, as their location affects the hyperbola's shape and the way it opens.

Asymptotes of a Hyperbola

Asymptotes are straight lines that approach the hyperbola but never intersect it. They guide the curvature of the hyperbola and predict its direction as it extends towards infinity.
In the standard form equation, the asymptotes for our hyperbola \(\frac{y^2}{9} - \frac{x^2}{25} = 1\) are given by the formula \(y = \pm \frac{a}{b}x\).

• Plugging in the known values, \(a = 3\) and \(b = 5\), these become \(y = \pm \frac{3}{5}x\).

These asymptotes cross at the origin and provide a boundary framework that the hyperbola approaches but never touches. When sketching, you align your curves to tend toward these lines, forming the hyperbola's general pathway.