Problem 11 - Precalculus with Calculus Previews

Question

In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ \frac{(y-4)^{2}}{36}-x^{2}=1 $$

Step-by-Step Solution

Answer

Center: (0, 4); Vertices: (0, 10), (0, -2); Foci: (0, 4±√37); Asymptotes: y=6x+4, y=-6x+4; Eccentricity: √37/6.

Step 1: Identify the type of conic section

The equation is of the form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), which represents a vertical hyperbola centered at \((h, k)\).

Step 2: Determine the center

For the equation \( \frac{(y-4)^2}{36} - x^2 = 1 \), the center \((h, k)\) is found directly from the expression in the numerators.Center: \((0, 4)\).

Step 3: Identify the vertices

For a vertical hyperbola \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), the vertices are at \((h, k \pm a)\).Given \(a^2 = 36\), thus \(a = 6\).Vertices: \((0, 4 \pm 6) = (0, 10) \) and \((0, -2)\).

Step 4: Calculate the foci

For a vertical hyperbola, the foci are at \((h, k \pm c)\), where \(c = \sqrt{a^2 + b^2}\).Given \(b^2 = 1\), thus \(b = 1\).Calculate \(c\):\[c = \sqrt{36 + 1} = \sqrt{37}.\]Foci: \((0, 4 \pm \sqrt{37})\).

Step 5: Find the equations of the asymptotes

The asymptotes for a vertical hyperbola \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) are given by:\[ y - k = \pm \frac{a}{b}(x-h).\]Since \(a = 6\) and \(b = 1\), substitute and simplify:\[y - 4 = \pm 6x.\]So the asymptotes are \( y = 6x + 4 \) and \( y = -6x + 4 \).

Step 6: Compute the eccentricity

Eccentricity \(e\) for a hyperbola is given by:\[e = \frac{c}{a}.\]Substitute the values of \(c\) and \(a\):\[e = \frac{\sqrt{37}}{6}.\]Thus, \(e = \frac{\sqrt{37}}{6}.\)

Step 7: Graph the hyperbola

Graphing involves plotting the center \((0, 4)\), vertices \((0, 10)\) and \((0, -2)\), and asymptotes \(y = 6x + 4 \) and \(y = -6x + 4 \). Sketch the hyperbola opening vertically along the lines of the asymptotes.

Key Concepts

Conic Sections

Conic sections are magnificent shapes that arise from slicing a cone with a plane at different angles. They include circles, ellipses, parabolas, and hyperbolas. In particular, a hyperbola is one such conic section. It features two distinct branches and is formed when the plane cuts through both halves of the cone. In our given exercise, the hyperbola is vertical. This is evident because the equation takes the form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) with the \( y \) term first, signaling a vertical orientation.

Understanding conic sections is crucial for recognizing and differentiating these shapes when solving mathematical problems. Hyperbolas, like other conic sections, have a center, foci, vertices, and asymptotes, which define their overall shape.

Eccentricity

Eccentricity is an important concept in understanding conic sections. It measures how much a conic section deviates from being circular. Specifically, for hyperbolas, the eccentricity value is always greater than 1.

For hyperbolas, eccentricity \( e \) is calculated using the formula \( e = \frac{c}{a} \), where \( c \) is the distance from the center to each focus, and \( a \) is the distance from the center to each vertex.

• In our given exercise, we calculated \( c \) as \( \sqrt{37} \), and \( a \) is 6, therefore the eccentricity is \( e = \frac{\sqrt{37}}{6} \).

This ratio gives us insight into the hyperbola’s shape. The further the foci are from the center, the more elongated the hyperbola. This makes the concept of eccentricity vital for understanding the specific nature of hyperbolas and other conics.

Asymptotes

Asymptotes are lines that a curve approaches but never actually reaches. They provide a framework that guides the overall direction of a hyperbola’s branches. In a hyperbola, these lines pass through the hyperbola's center and define the slope of its open ends.

The asymptotes for a hyperbola with a vertical orientation are derived from the formula \( y - k = \pm \frac{a}{b}(x-h) \). Here, \( k \) is the \( y \)-coordinate of the center, \( h \) is the \( x \)-coordinate, \( a \) is the distance from the center to a vertex along the transverse axis, and \( b \) is related to the conjugate axis.

• In our exercise, \( a = 6 \) and \( b = 1 \), leading to the asymptote equations \( y = 6x + 4 \) and \( y = -6x + 4 \).

Understanding these asymptotes allows us to sketch the hyperbola accurately, as they indicate the direction in which the hyperbola opens.