Problem 27

Question

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$\frac{(y+6)^{2}}{1 / 9}-\frac{(x-2)^{2}}{1 / 4}=1$$

Step-by-Step Solution

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Answer

The center is (2, -6) with a radius along x and y of 0.5 and 1/3 respectively. Vertices are at (2, -20/3) and (2, -16/3). The foci are at (2, -6.55) and (2, -5.45). The equations for the asymptotes are y = 2x - 10 and y = -2x + 2.

1Step 1: Identify the center

The center of the hyperbola can be found by identifying the values that shift the hyperbola away from the origin on the x and y axes. This includes the values subtracted from x and added to y. The center is given by $(h, k)$, so in our case, we have $h = 2$ and $k = -6$, hence the center is $(2, -6)$.

2Step 2: Identify the radius along the x-axis (a) and along the y-axis (b)

The denominators under each square in the equation represent the square of the distance from the center to the vertices along each axis. The square root of each denominator provides this distance value. In our case, the square root of $\frac{1}{4}$ which is 0.5 and the square root of $\frac{1}{9}$ which is $\frac{1}{3}$. The distances represent the lengths along the x-axis and y-axis respectively. However, in this case, as the y term is positive and the x term is negative, the y-axis length represents 'a' and the x-axis length represents 'b'. So we have $a = \frac{1}{3}$ and $b = 0.5$.

3Step 3: Find the vertices

The vertices are given by $(h, k \pm a)$. So, the vertices are $(2, -6 \pm \frac{1}{3}) = (2, -\frac{20}{3}), (2, -\frac{16}{3})$

4Step 4: Calculate the foci

The foci are found by first finding the distance from the center to the foci, represented by 'c'. This is found using the relationship $c = \sqrt{a^2 + b^2}$. In our case, $c = \sqrt{(\frac{1}{3})^2 + 0.5^2} = 0.55$. Then, as our a is the y-axis, the foci are at $(h, k \pm c)$ which is $(2, -6 \pm 0.55) = (2, -6.55), (2, -5.45)$

5Step 5: Find the equations of the asymptotes

The equations for the asymptotes of a vertical hyperbola are given by $k \pm \frac{a}{b} (x - h) = y$. So, substituting our h, k, a and b we get $-6 \pm \frac{1/3}{0.5} (x - 2) = y$, which simplifies into two equations $y = 2x - 10$ and $y = -2x + 2$

6Step 6: Draw the hyperbola

Using all this obtained information, plot points for the center, vertices and foci on a graph. Draw lines representing the asymptotes. The hyperbola should approach these lines but never cross them. Positions of the vertices and foci and center will provide a generate outlook of the shape of hyperbola.

Key Concepts

Center of the HyperbolaVertices of the HyperbolaFoci of the HyperbolaAsymptotes of the Hyperbola

Center of the Hyperbola

The center of a hyperbola is an important feature as it serves as the point from which the hyperbola is defined. To find the center, we look at the values associated with the variables in the equation. For an equation of the form $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$, the center is

$(h, k)$

In the given hyperbola equation $\frac{(y+6)^{2}}{1 / 9}-\frac{(x-2)^{2}}{1 / 4}=1$, the values $y+6$ and $x-2$ suggest the shifts from the origin. The center is found by isolating the changes in these terms, giving us the coordinates:

$h = 2$
$k = -6$

Thus, the center of the hyperbola is at the point $(2, -6)$. This spot is crucial, as it acts as the balancing point of the hyperbola shape.

Vertices of the Hyperbola

The vertices of a hyperbola reveal the most extreme points along its axis and are determined from the center and the distance 'a'. For a vertical hyperbola, these vertices are located at

$(h, k \pm a)$

In our equation, the term under $(y+6)$ is used to find 'a'. This gives us:

$a = \sqrt{\frac{1}{9}} = \frac{1}{3}$

Substituting these values provides:

First Vertex: $(2, -6 + \frac{1}{3}) = (2, -\frac{16}{3})$
Second Vertex: $(2, -6 - \frac{1}{3}) = (2, -\frac{20}{3})$

These vertices mark the endpoints of the hyperbola as it stretches vertically.

Foci of the Hyperbola

The foci are specific points that lie along the axis of symmetry of a hyperbola. They help define its shape and are a fixed distance, 'c', from the center. The formula for 'c' relies on 'a' and 'b' as:

$c = \sqrt{a^2 + b^2}$

Given:

$a = \frac{1}{3}$
$b = 0.5$

We calculate:

$c = \sqrt{\left(\frac{1}{3}\right)^2 + 0.5^2} = 0.55$

Positioning the foci for the vertical hyperbola involves computation as:

First Focus: $(2, -6 + 0.55) = (2, -5.45)$
Second Focus: $(2, -6 - 0.55) = (2, -6.55)$

These foci indicate the pronounced inward curvature toward these points within the hyperbola.

Asymptotes of the Hyperbola

Asymptotes are crucial lines that hyperbolas approach but never touch. They aid in sketching the overall curve. For a vertical hyperbola, asymptotes have the form: