Problem 10 - Precalculus with Calculus Previews

Question

In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ \frac{(x+2)^{2}}{10}-\frac{(y+4)^{2}}{25}=1 $$

Step-by-Step Solution

Answer

Center: (-2, -4). Vertices: (-2±√10, -4). Foci: (-2±√35, -4). Asymptotes: y+4=±√10/2(x+2). Eccentricity: √3.5.

Step 1: Identify the Standard Form

The equation of a hyperbola is given as \( \frac{(x+2)^{2}}{10} - \frac{(y+4)^{2}}{25} = 1 \). This matches the standard form of a hyperbola \( \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1 \). Comparing these, we identify: \((h, k) = (-2, -4)\), \(a^{2} = 10\), and \(b^{2} = 25\).

Step 2: Find the Center

The center of the hyperbola is directly given by \((h, k)\). Thus, the center is \((-2, -4)\).

Step 3: Determine the Vertices

For the horizontal hyperbola, the vertices are \((h \pm a, k)\). First, find \(a\) using \(a^{2} = 10\), so \(a = \sqrt{10}\). Thus, the vertices are \((-2 \pm \sqrt{10}, -4)\).

Step 4: Calculate the Foci

The foci of a hyperbola are given by \((h \pm c, k)\), where \(c^{2} = a^{2} + b^{2}\). Compute \(c^{2} = 10 + 25 = 35\), giving \(c = \sqrt{35}\). Thus, the foci are \((-2 \pm \sqrt{35}, -4)\).

Step 5: Equation of Asymptotes

The asymptotes for a horizontal hyperbola are given by \(y = k \pm \frac{b}{a}(x-h)\). Here, \(\frac{b}{a} = \frac{5}{\sqrt{10}} = \frac{\sqrt{10}}{2}\). Hence, the asymptotes are \(y + 4 = \pm \frac{\sqrt{10}}{2}(x + 2)\).

Step 6: Compute Eccentricity

The eccentricity \(e\) is defined as \(e = \frac{c}{a}\). We have \(c = \sqrt{35}\) and \(a = \sqrt{10}\), so \(e = \frac{\sqrt{35}}{\sqrt{10}}\). Simplify this to \(e = \sqrt{\frac{35}{10}} = \sqrt{3.5}\).

Step 7: Graph the Hyperbola

To graph, start by plotting the center \((-2,-4)\). From the center, plot the vertices \((-2+\sqrt{10}, -4)\) and \((-2-\sqrt{10}, -4)\). Draw the asymptotes \(y+4 = \pm \frac{\sqrt{10}}{2}(x+2)\). Sketch the hyperbola opening left and right, approaching these asymptotes.

Key Concepts

Center of Hyperbola

The center of a hyperbola is the point around which the entire figure is balanced. It is a crucial reference point for plotting the hyperbola and calculating other key features. In the given equation of the hyperbola, \[ \frac{(x+2)^{2}}{10} - \frac{(y+4)^{2}}{25} = 1, \]we can discern the center by recognizing the standard form, which is \[ \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1. \]Comparing these, we identify \( h = -2 \) and \( k = -4 \).

• The center is therefore located at \((-2, -4)\).
• This point serves as the origin for measuring distances to vertices and foci.

Vertices of Hyperbola

Vertices of a hyperbola are the points where the hyperbola crosses its transverse axis. For a horizontal hyperbola, these points can be found using the formula \((h \pm a, k)\).

• First, we calculate \(a\) where \(a^{2} = 10\).
  This gives us \(a = \sqrt{10}\).
• The vertices of this hyperbola are therefore: \((-2 + \sqrt{10}, -4)\) and \((-2 - \sqrt{10}, -4)\).

Vertices are essential as they mark the closest distance the curves of the hyperbola approach the center along the transverse axis.

Foci of Hyperbola

Foci are points located along the transverse axis of the hyperbola that determine the hyperbola's shape. The distance from the center to these points affects the eccentricity or the "stretch" of the hyperbola. To find the foci, use the formula \((h \pm c, k)\) where \(c^{2} = a^{2} + b^{2}\).

• We know \(a^{2} = 10\) and \(b^{2} = 25\), so \(c^{2} = 35\).
• Solving for \(c\), we find \(c = \sqrt{35}\).
• The foci are located at \((-2 + \sqrt{35}, -4)\) and \((-2 - \sqrt{35}, -4)\).

These foci demonstrate a distinct property: as you sum the distances from any point on the hyperbola to each focus, the result is always constant.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never meets. They play a vital role in sketching the shape of the hyperbola. For a hyperbola in the standard horizontal form, the equations of the asymptotes are given by \(y = k \pm \frac{b}{a}(x-h)\).

• First, calculate \(\frac{b}{a}\) where \(b = 5\) and \(a = \sqrt{10}\).
• This gives us \(\frac{b}{a} = \frac{5}{\sqrt{10}} = \frac{\sqrt{10}}{2}\).
• The asymptote equations are therefore:
  \(y + 4 = \pm \frac{\sqrt{10}}{2}(x + 2)\).

These asymptotes guide the direction and steepness of the hyperbola branches, helping in predicting where they will extend.

Eccentricity of Hyperbola

Eccentricity is a measure of how much the shape of the hyperbola deviates from being circular, or how "stretched" it is. It is calculated using the formula \(e = \frac{c}{a}\).

• Here, we calculated that \(c = \sqrt{35}\) and \(a = \sqrt{10}\).
• The eccentricity \(e\) is hence \(\frac{\sqrt{35}}{\sqrt{10}}\).
• This simplifies to \(e = \sqrt{3.5}\).

The eccentricity value tells us that the hyperbola is more elongated relative to its transverse axis.