Problem 6 - Precalculus with Calculus Previews

Question

In Problems \(1-20\), find the center, foci, vertices, asymptotes, and eccentricity of the given hyperbola. Graph the hyperbola. $$ 5 x^{2}-5 y^{2}=25 $$

Step-by-Step Solution

Answer

Center: (0, 0); Vertices: (±√5, 0); Foci: (±√10, 0); Asymptotes: y = ±x; Eccentricity: √2.

Step 1: Simplify the Equation

Start with the given equation of the hyperbola: \(5x^2 - 5y^2 = 25\). Divide every term by 25 to get the equation in standard form:\[\frac{x^2}{5} - \frac{y^2}{5} = 1\] This represents a hyperbola centered at the origin with the transverse axis along the x-axis.

Step 2: Identify the Center

The standard form of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) reveals that the center of the hyperbola is at the origin: \((0, 0)\).

Step 3: Find the Vertices

For a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices are \((\pm a, 0)\). Here, \(a^2 = 5\), so \(a = \sqrt{5}\). The vertices are at \((\pm \sqrt{5}, 0)\).

Step 4: Find the Foci

The foci of the hyperbola are given by \((\pm c, 0)\), where \(c = \sqrt{a^2 + b^2}\). Since \(b^2 = 5\), calculate\[c = \sqrt{5 + 5} = \sqrt{10}\]Thus, the foci are located at \((\pm \sqrt{10}, 0)\).

Step 5: Determine Asymptotes

For the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations of the asymptotes are given by \(y = \pm \frac{b}{a}x\). Since \(b = \sqrt{5}\) and \(a = \sqrt{5}\), the asymptotes are:\[y = \pm x\]

Step 6: Calculate Eccentricity

The eccentricity \(e\) of a hyperbola is calculated using \(e = \frac{c}{a}\). Here, \(a = \sqrt{5}\) and \(c = \sqrt{10}\), so:\[e = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}\]

Step 7: Graph the Hyperbola

Draw the center at (0,0). Plot the vertices at \((\pm \sqrt{5}, 0)\) and foci at \((\pm \sqrt{10}, 0)\). Sketch the asymptotes \(y = x\) and \(y = -x\). Draw the hyperbola opening left and right, approaching the asymptotes.

Key Concepts

Understanding Eccentricity

Eccentricity is a measure of how "stretched out" a conic section is compared to a circle. For hyperbolas, it tells us how much the conic deviates from being circular.

• If the eccentricity is 1, it's a parabola.
• If greater than 1, it's a hyperbola.
• If less than 1, it's an ellipse.

In the hyperbola equation, we find eccentricity using the formula:\[ e = \frac{c}{a} \]where \( c \) is the distance from the center to the foci and \( a \) is the distance from the center to the vertices. In our example, with \( a = \sqrt{5} \) and \( c = \sqrt{10} \), we calculate:\[ e = \sqrt{2} \]A higher value of eccentricity like \( \sqrt{2} \) signifies the hyperbola is more elongated.

Decoding Asymptotes

In hyperbolas, asymptotes are lines that the curve approaches but never touches, providing a visual boundary for the arms of the hyperbola.These lines help to visualize the direction in which the hyperbola's branches open.For the standard form hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the asymptotes are given by:\[ y = \pm \frac{b}{a}x \]In our hyperbola, since both \( a \) and \( b \) are \( \sqrt{5} \), the asymptotes simplify to:\[ y = \pm x \]You can draw these lines on the graph by using the slopes \( 1 \) and \(-1\), creating guidelines for the hyperbola.

Identifying Vertices

Vertices are crucial points on a hyperbola where each branch turns its sharpest direction.For a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), vertices are located at:\[ (\pm a, 0) \]In our example, we have:\[ a = \sqrt{5} \]Thus, the vertices are at:\[ (\pm \sqrt{5}, 0) \]These points, along with the center, help in sketching the basic shape of the hyperbola.

Locating the Foci

The foci (plural of focus) of a hyperbola are two specific points used to construct and define the curve. They play a role in maintaining the constant difference in distances from any point on the hyperbola to each focus.For the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the foci are located at:\[ (\pm c, 0) \]where\[ c = \sqrt{a^2 + b^2} \]In our hyperbola, \( a^2 = 5 \) and \( b^2 = 5 \), making:\[ c = \sqrt{10} \]Thus, the foci are:\[ (\pm \sqrt{10}, 0) \]These points help in shaping the curve, as every hyperbola is defined by its foci.