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Chapter 7: Problem 6

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

Short Answer

Expert verified
Center: (0, 0); Vertices: (卤鈭�5, 0); Foci: (卤鈭�10, 0); Asymptotes: y = 卤x; Eccentricity: 鈭�2.

Step by step solution

01

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.
02

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: \((0, 0)\).
03

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 \((\pm \sqrt{5}, 0)\).
04

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 \((\pm \sqrt{10}, 0)\).
05

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:\[y = \pm x\]
06

Calculate Eccentricity

The eccentricity \(e\) of a hyperbola is calculated using \(e = \frac{c}{a}\). Here, \(a = \sqrt{5}\) and \(c = \sqrt{10}\), so:\[e = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}\]
07

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Describe geometrically all points in 3-space whose coordinates satisfy the given condition(s). $$ x^{2}+y^{2}+(z-1)^{2}=4, z=2 $$

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Center \((2,4),\) one vertex \((2,5),\) one asymptote \(2 y\) \(-x-6=0\)

Conjugate Hyperbolas The hyperbolas $$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \quad \text { and } \quad \frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1 $$ are said to be conjugates of each other. (a) Find an equation of the hyperbola that is conjugate to $$ \frac{x^{2}}{25}-\frac{y^{2}}{144}=1 $$ (b) Discuss how the graphs of conjugate hyperbolas are related.

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Foci \((0,\pm 3),\) one vertex \(\left(0,-\frac{3}{2}\right)\)

Find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ x^{2}+3 y^{2}+18 y+18=0 $$
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