Problem 36 - College Algebra

Question

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes \(\frac{(x+2)^{2}}{9}-\frac{y^{2}}{25}=1\)

Step-by-Step Solution

Answer

The center of the hyperbola is at (-2,0), the vertices are at (-5,0) and (1,0) and the foci are at (-2±sqrt(34),0). The equation of the asymptotes are \( y = ± (\frac{5}{3})(x+2) \). By plotting these elements on the graph, the shape and direction of the hyperbola can be established.

Step 1: Identify the center

The center (h, k) of the hyperbola can be found from the equation \(\frac{(x+2)^{2}}{9}-\frac{y^{2}}{25}=1\). Here, the center is at (-2, 0).

Step 2: Find the vertices

The vertices are at a distance a from the center along the x-axis in this case. Since \(\frac{(x+2)^{2}}{9}=1\) when y=0, 'a' will be square root of 9, which is 3. So, the vertices are at (-2±3, 0) which means (-5,0) and (1,0).

Step 3: Find the foci

The foci are at a distance c from the center. Using formula c = sqrt(a^2 + b^2), where b is the square root of 25, we find that c = sqrt(9 + 25) = sqrt(34). Therefore, the foci are at (-2±sqrt(34),0).

Step 4: Find the equations of the asymptotes

The asymptotes of the hyperbola are given by the equation \( y = k ± (\frac{b}{a})(x-h) \), where the slope is given by \( \pm \frac{b}{a} \). So, the asymptotes of the given hyperbola are \( y = ± (\frac{5}{3})(x+2) \).

Step 5: Graph the hyperbola

Draw the center, vertices and foci on the x-axis. Then draw the asymptotes, which are diagonals of the rectangle that can be drawn with sides of 2a and 2b. The hyperbola will approach these asymptotes. The graph of hyperbola opens right and left through the vertices because the positive term in the equation is \(x^{2}\).

Key Concepts

Equation of a Hyperbola

A hyperbola is a type of conic section that looks somewhat like two mirrored perpendicular parabolas. The standard form of a hyperbola's equation can appear in two ways depending on its orientation. For a hyperbola that opens horizontally, the equation is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). For a hyperbola that opens vertically, the roles of \(x\) and \(y\) swap.
In the equation \( \frac{(x+2)^2}{9} - \frac{y^2}{25} = 1 \), we can see it takes the format for opening horizontally:

• \(h = -2\) and \(k = 0\), setting the center at \((-2,0)\).
• \(a^2 = 9\) gives \(a = 3\). This is the distance from the center to each vertex along the x-axis.
• \(b^2 = 25\) results in \(b = 5\). This indicates that, although a hyperbola does not have a physical width like an ellipse, setting up the box to draw asymptotes needs this measure.

Recognizing these components lets us interpret information about the shape and orientation of the hyperbola and is crucial for graphing it.

Asymptotes of a Hyperbola

Asymptotes are the lines that the hyperbola approaches but never quite touches. They guide the framework of the hyperbola and are a vital element to understand its graphical shape. For a horizontal hyperbola described by \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \], the asymptotes are given by:
\[ y = k \pm \left(\frac{b}{a} \right)(x-h) \]

• In our example, \( a = 3 \), \( b = 5 \), \( h = -2 \), \( k = 0 \).
• Plugging these into the equation gives the asymptotes: \( y = \pm \frac{5}{3}(x + 2) \).
• These lines are not part of the hyperbola itself but are crucial guides for drawing it accurately.

These mathematical guidelines help position the hyperbola when sketching it on a graph and indicate its broad divergence direction.

Foci of a Hyperbola

Foci are special points of a hyperbola from which the distances are measured to define the curve. Every point on the hyperbola maintains a constant distance difference from these foci. This design sets a hyperbola apart from other conic sections like ellipses or circles.
To locate the foci:

• Utilize the formula \( c = \sqrt{a^2 + b^2} \).
• In our equation, \( a = 3 \) and \( b = 5 \), we derive \( c = \sqrt{9 + 25} = \sqrt{34} \).
• This interval \( c \) determines how far each focus lies from the center along the x-axis, making the foci \((-2 \pm \sqrt{34}, 0)\).

Knowing where the foci is essential for comprehensive understanding and proper graphing of the hyperbola, as they further emphasize the curve's geometry and balance.