Problem 38 - Algebra and Trigonometry

Question

use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$ \frac{(y-2)^{2}}{36}-\frac{(x+1)^{2}}{49}=1 $$

Step-by-Step Solution

Answer

The center of the hyperbola is (-1,2), the vertices are (-1,8) and (-1,-4). The equations of the asymptotes are \(y = 2 + 6(x + 1)\) and \(y = 2 - 6(x + 1)\). The foci of the hyperbola are located at (-1, 2+\sqrt{85}) and (-1, 2-\sqrt{85}).

Step 1: Identify the Center

The center of the hyperbola is given by \((h, k)\). It is the center point from which the distance to the vertices and the foci is measured. The standard equation of the hyperbola is given by \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\). By comparing this formula with the given equation \(\frac{(y-2)^2}{36} - \frac{(x+1)^2}{49} =1\), we see that the center is located at \((-1, 2)\).

Step 2: Determine Vertices and Asymptotes

Upon observing the given equation, the value beneath \(y-2)^2\) corresponds to \(a^2 = 36\), hence \(a = 6\). The vertices are located at an 'a' distance from the center in the direction of the major axis of the hyperbola. Hence, vertices are located at (-1, 2+6) i.e. (-1, 8) and (-1, 2-6) i.e. (-1, -4). The equation of asymptotes will be in the form of \(\frac{(y-k)^2}{a^2} = \pm \frac{(x-h)^2}{b^2}\). Hence, equations for asymptotes are \(y = 2 + 6(x + 1)\) and \(y = 2 - 6(x + 1)\).

Step 3: Find the Foci

The foci are located at a distance of 'c' from the center. Here, \(c^2 = a^2 + b^2 = 36 + 49 = 85\), hence \(c = \sqrt{85}\). Therefore, the foci are located at (-1, 2+\sqrt{85}) and (-1, 2-\sqrt{85}).

Key Concepts

Center of Hyperbola

In mathematics, hyperbolas are intriguing shapes that have specific characteristics. The center of a hyperbola is its balancing point. It's the spot where the two branches of the hyperbola are symmetrically distributed. For the equation \[\frac{(y-2)^2}{36} - \frac{(x+1)^2}{49} = 1\]you can identify the center by comparing it to the standard hyperbola equation: \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1.\]Here, \(h\) and \(k\) correspond to the values that shift the hyperbola in the Cartesian plane. In this equation, the center is \((-1, 2)\). This point is crucial because it helps to determine other key elements, such as the vertices and foci.
When you think of a hyperbola, always start at its center, as this coordinates everything else.

Vertices of Hyperbola

Vertices are important landmarks on a hyperbola. They are the points where each branch is closest to the center. For a vertical hyperbola like this one:\[\frac{(y-2)^2}{36} - \frac{(x+1)^2}{49} = 1,\]the vertices are aligned with the center along the y-axis.
In this case, the given equation reveals that \(a^2 = 36\), so \(a = 6\). This means the vertices are a distance \(a\) from the center, specifically \(a\) units up and \(a\) units down the y-axis.

• The upper vertex is at \((-1, 8)\).
• The lower vertex is at \((-1, -4)\).

Vertices help define the hyperbola's shape, being essential for sketching or understanding its geometric properties.

Asymptotes of Hyperbola

Asymptotes are the invisible lines that guide the shape of a hyperbola's branches and they intersect at the center. They are not part of the hyperbola itself but point to how the branches curve outward. For our given hyperbola:\[\frac{(y-2)^2}{36} - \frac{(x+1)^2}{49} = 1,\]the equations for the asymptotes can be calculated using \[y - k = \pm \frac{a}{b}(x - h).\]Using \(a = 6\) and \(b = 7\), the equations become:- \(y = 2 + \frac{6}{7}(x + 1)\)- \(y = 2 - \frac{6}{7}(x + 1)\)
These lines tell you that as you move further from the center on the hyperbola, the branches will look more like straight lines approaching these asymptotes. While they seem simple, understanding asymptotes allows you to predict the hyperbola’s behavior at extremes.

Foci of Hyperbola

The foci (plural for focus) are distinct points of interest in a hyperbola, showing how it 'opens up'. For our equation:\[\frac{(y-2)^2}{36} - \frac{(x+1)^2}{49} = 1,\]we find the foci using the formula \(c^2 = a^2 + b^2\). With \(a^2 = 36\) and \(b^2 = 49\), we find that \(c^2 = 85\) and thus \(c = \sqrt{85}\).
Consequently, the foci lie along the y-axis centered around the center point:

• The upper focus is at \((-1, 2 + \sqrt{85})\).
• The lower focus is at \((-1, 2 - \sqrt{85})\).

The foci are significant because they can be used to construct a hyperbola and understand how it stretches or compresses. Knowing where the foci are can also help in various physical applications, as they often correspond to points of concentrated energy or signal in real-world phenomena.