Problem 44 - Algebra 2

Question

Find the vertices and the asymptotes of each hyperbola. $$ 9 y^{2}-36 x^{2}=144 $$

Step-by-Step Solution

Answer

Vertices are at (0, -4) and (0, 4) and Asymptotes have equations \(y = 2x\) and \(y = -2x\).

Step 1: Get the standard form of the equation

To get the standard form of the equation, divide the given equation by 144. So, the equation becomes \(\frac{y^2}{16} - \frac{x^2}{4} = 1\).

Step 2: Identify the center, a and b

From the standard form, we see that the center of the hyperbola is at the origin, (0,0), since h and k are both 0. The value of \(a\) is 4 and \(b\) is 2 because in the standard form of equation of hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), \(a^2\) equals 16 and \(b^2\) equals 4.

Step 3: Find the vertices

The vertices of a hyperbola are a distance 'a' from the center. As the major axis is vertical in this case, the vertices are at (0,±4).

Step 4: Find the asymptotes

The equations of the asymptotes in this case are y = ±(a/b)x. The equations of the asymptotes are therefore \(y = ±2x\).

Key Concepts

Standard Form of a Hyperbola

The standard form of a hyperbola helps to identify the shape and orientation of the curve. For any hyperbola, the equation can take two forms:

• Horizontal hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• Vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

The difference lies in which term comes first. If \(y\) is first, the hyperbola opens upward and downward, indicating a vertical hyperbola. When \(x\) is first, the hyperbola opens left and right, indicating a horizontal hyperbola. After transforming the equation \(9y^2 - 36x^2 = 144\) by dividing it by 144, we get \(\frac{y^2}{16} - \frac{x^2}{4} = 1\). This form tells us we are dealing with a vertical hyperbola.

Vertices of a Hyperbola

Vertices are key components of a hyperbola, representing the points where the curve is closest to its center. For a hyperbola given by the standard form, vertices are found at a distance \(a\) from the center along the major axis. In the equation \(\frac{y^2}{16} - \frac{x^2}{4} = 1\), we identify \(a^2\) as 16, thus \(a\) as 4. The center of our hyperbola is at \((0,0)\), leading us to find the vertices at \((0, +4)\) and \((0, -4)\). These vertices tell us that the hyperbola stretches vertically along the \(y\)-axis.

Asymptotes of a Hyperbola

Asymptotes are lines that a hyperbola approaches but never actually reaches. For hyperbolas, they provide a way to sketch the curve correctly by showing the shape's extent and direction. The equations for asymptotes in our vertical hyperbola \(\frac{y^2}{16} - \frac{x^2}{4} = 1\) are derived from the slope \(\pm \frac{a}{b}\). Substituting \(a = 4\) and \(b = 2\), we obtain the asymptotes as \(y = \pm 2x\). This means the hyperbola will stretch outwards to match the angle formed by these paths.

Conic Sections

Conic sections are curves obtained by intersecting a plane with a cone. Depending on the angle and position of the intersection, they can form different shapes such as circles, ellipses, parabolas, and hyperbolas. Hyperbolas arise when a plane slices through both halves of a double cone, creating two mirrored curves. They have two characteristic branches, which are symmetrical and extend indefinitely. Understanding the properties of a hyperbola, like its shape, vertices, and asymptotes, is essential to mastering conic sections. Each type of conic section has unique properties and applications in various fields, from architecture to astronomy.