Problem 37 - Precalculus

Question

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Use a graphing utility to graph the hyperbola and its asymptotes. $$9 y^{2}-x^{2}+2 x+54 y+62=0$$

Step-by-Step Solution

Answer

The center of the hyperbola is at (1, -3). Its vertices are at (1, -3±√2), co-vertices at (1±√18, -3), and foci at (1, -3±2√5). The equations of the asymptotes are \(y+3=±(√{2}/√{18})(x-1)\).

Step 1: Complete the square

Rearrange and complete the square for x and y terms to rewrite the equation into the usual standard hyperbola form. Group x's and y's together: \[9y^{2} + 54y - x^{2} +2x = -62\]. Then, make the coefficient of y² and x² as 1 by adding 1 to four times the coefficient of y and x respectively, yielding: \[9(y^{2} + 6y + 9) - (x^{2} - 2x +1) = -62+81-1\]. So the equation becomes: \[9(y+3)^{2} - (x-1)^{2}=18\]

Step 2: Rewrite in standard form

Divide through by 18 to put the definition in the standard form: \[(y+3)^{2}/2 - (x-1)^{2}/18 =1\] Here, a^2 = 2, b^2 = 18

Step 3: Find the center, vertices, and co-vertices

From the standard form, we can identify the center of the hyperbola (h, k) = (1, -3). The vertices are found 'a' units above and below the center, so they're at (1, -3±√2). The co-vertices are 'b' units left and right of the center, so they're at (1±√18, -3)

Step 4: Find the foci and asymptotes

The foci are 'c' units above and below the center, where \(c=sqrt{a^2 + b^2}\). So, \( c = sqrt{2+18} = sqrt{20} = 2√5\), and the foci are at (1, -3±2√5). The asymptotes are given by the equations \(y-k = ±(a/b)(x-h)\), or \(y+3=±(√{2}/√{18})(x-1)\)

Key Concepts

Center of Hyperbola

The center of a hyperbola is crucial in understanding its geometry. It acts as a point from which the hyperbola's symmetry reflects. For a given hyperbola formula, the center can be easily identified in the standard equation format. This format is:\[(x - h)^2/a^2 - (y - k)^2/b^2 = 1\] \[\text{or} \] \[(y - k)^2/a^2 - (x - h)^2/b^2 = 1\]. Here, \((h, k)\) represent the center coordinates of the hyperbola.
In the problem at hand, the equation \[(y+3)^2/2 - (x-1)^2/18 = 1\] points out the center as \((1, -3)\). This means the hyperbola is centered at \(x = 1\) and \(y = -3\). The concept of the center helps us establish a reference for finding other important parts of the hyperbola, such as vertices, foci, and asymptotes.

Vertices of Hyperbola

The vertices of a hyperbola are the points where each branch of the hyperbola is closest to the center. They lie along the transverse axis, the axis extending through the center, touching both vertices. In the standard form equation, \[(y+3)^2/2 - (x-1)^2/18 = 1\], we identify "a" to help find these vertices. Here \(a^2 = 2\), so \(a = \sqrt{2}\).
Given the center of the hyperbola at \((1, -3)\), we find the vertices by moving \(a\) units along the transverse axis from the center. Since this hyperbola opens vertically, moving along \(y\) gives the vertices at \((1, -3 + \sqrt{2})\) and \((1, -3 - \sqrt{2})\). These points help define the shape and direction of the branches of the hyperbola.

• Recall the transverse axis tells us where the branches of the hyperbola are situated.
• The length between the vertices is \(2a\).

Foci of Hyperbola

The foci of the hyperbola provide essential information regarding its shape and are always found further from the center than the vertices. They lie along the same axis as the vertices. To find the foci, we use the formula \(c = \sqrt{a^2 + b^2}\) where \(c\) represents the distance from the center to each focus. In our example: \[a^2 = 2, \quad b^2 = 18\]\[c = \sqrt{2 + 18} = \sqrt{20} = 2\sqrt{5}\]
Thus, given the center at \((1, -3)\), the foci are \((1, -3 + 2\sqrt{5})\) and \((1, -3 - 2\sqrt{5})\). The foci help determine the "spread" of the hyperbola branches. The closer the foci, the flatter the hyperbola; the further apart they are, the more open the hyperbola appears.

Equations of Asymptotes

Asymptotes of a hyperbola are lines that the branches approach but never touch. They give a guideline of the hyperbola's growth. In the standard equation, the asymptotes can be found using the formula:
\(y - k = \pm \frac{a}{b}(x - h)\). For the provided problem, we calculate the asymptotes using:\[a = \sqrt{2}\quad \text{and}\quad b = \sqrt{18}\]\[y + 3 = \pm \frac{\sqrt{2}}{\sqrt{18}}(x - 1)\]
Upon simplification, this becomes:\[y + 3 = \pm \frac{1}{3}(x - 1)\]
These linear equations outline the directions along which the hyperbola branches extend. The asymptotes cross at the center of the hyperbola, acting as a cue for the trailing edges of the branches as they extend infinitely closer but never intersecting the asymptotes.