Problem 43
Question
In Exercises 43-50, convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$x^{2}-y^{2}-2 x-4 y-4=0$$
Step-by-Step Solution
Verified Answer
The equation in standard form is \((x-1)^2 -(y+2)^2 = 1\). The coordinates of the foci are \((1+√2, -2)\) and \((1-√2, -2)\), and the equations of the asymptotes are \(y = x - 3\) and \(y = -x + 1\).
1Step 1: Convert to Standard Form
Firstly, rearrange terms to bring the equation into a recognizable 'completing the square' form by pairing \(x\) terms and \(y\) terms together: \(x^{2}-2x-y^{2}+4y=-4.\) Next, complete the square on \(x\) and \(y\) terms. Half of 2 is 1 and \(1^2=1\), add this to both sides for the \(x\) terms. Half of 4 is 2 and \(2^2=4\), subtract this from both sides for the \(y\) terms. This gives us: \((x-1)^2-(y+2)^2=1 \). This is the equation in the standard form.
2Step 2: Graph the hyperbola
The center is at (1, -2), represented by the terms (h, k) in the standard form. Taking the square root of the coefficients of \(x\) and \(y\) gives the values of \(a\) and \(b\) (the transverse and conjugate axes half lengths). We get \(a=1\) and \(b=1\). This places vertices at (2, -2) and (0, -2), foci at (2 + √2, -2) and (2 - √2, -2), and asymptotes at \(y = (x - 1) ± 1\) as guidelines for sketching the hyperbola.
3Step 3: Find the Foci
The distance from the center to each focus point is given by \(c = \sqrt{a^2 + b^2}\). Substituting the found a and b values, we get \(c = \sqrt{1^2 + 1^2} = √2 \). So, the coordinates of the foci are \((1+√2, -2)\) and \((1-√2, -2)\)
4Step 4: Find the Asymptotes equations
The equations for the asymptotes of a hyperbola in the form \( (x-h)^2/a^2 - (y-k)^2 / b^2 = 1 \) is given by \(y = k ± (b/a)(x - h)\). Substituting the h, k, a and b values, we get: \(y = -2 ± (x - 1)\), which gives the equations of the asymptotes as \(y = x - 3\) and \(y = -x + 1\).
Key Concepts
HyperbolaStandard Form of HyperbolaAsymptotesFoci of Hyperbola
Hyperbola
A hyperbola is a type of conic section that appears as two open, mirror-image curves. It's different from other conic sections such as ellipses and circles. The key feature of a hyperbola is that for any point on it, the difference of the distances to two fixed points (the foci) is constant. This distinct property makes hyperbolas unique and often useful in various scientific applications, from orbits of celestial bodies to the shape of certain sound waves.
Hyperbolas can be horizontal or vertical in orientation, depending on the placement of the variable terms in their equations. When writing the equation of a hyperbola, the important components are its center, axes lengths, foci, and asymptotes, which help in sketching its graph and understanding its geometry.
Hyperbolas can be horizontal or vertical in orientation, depending on the placement of the variable terms in their equations. When writing the equation of a hyperbola, the important components are its center, axes lengths, foci, and asymptotes, which help in sketching its graph and understanding its geometry.
Standard Form of Hyperbola
The standard form of a hyperbola's equation allows us to easily identify and graph it. For horizontal hyperbolas, the equation is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] and for vertical hyperbolas, it switches to \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \].
In these equations:
In these equations:
- \((h, k)\) is the center of the hyperbola.
- \(a\) represents the distance from the center to each vertex along the transverse axis.
- \(b\) is the distance from the center to each vertex along the conjugate axis.
Asymptotes
Asymptotes are straight lines that the curves of a hyperbola approach but never actually meet or cross. They provide a skeleton for graphing the hyperbola. Asymptotes help in anticipating the shape and direction of the hyperbola branches.
The equations of the asymptotes for a hyperbola in standard form can be derived using the formula:\[ y = k \pm \frac{b}{a}(x - h) \]for horizontal hyperbolas, and \[ y = k \pm \frac{a}{b}(x - h) \]for vertical ones.
This formula uses the center \((h, k)\) and the values \(a\) and \(b\), which define the relative steepness of the asymptotes' slopes. Visualizing these guidelines helps ensure that the hyperbola's graph aligns correctly with its expected intersection points and paths.
The equations of the asymptotes for a hyperbola in standard form can be derived using the formula:\[ y = k \pm \frac{b}{a}(x - h) \]for horizontal hyperbolas, and \[ y = k \pm \frac{a}{b}(x - h) \]for vertical ones.
This formula uses the center \((h, k)\) and the values \(a\) and \(b\), which define the relative steepness of the asymptotes' slopes. Visualizing these guidelines helps ensure that the hyperbola's graph aligns correctly with its expected intersection points and paths.
Foci of Hyperbola
The foci of a hyperbola are two fixed points that are crucial in its definition and play a major role in its structure. The entire shape of a hyperbola pivots around these points. For any point on the hyperbola, the absolute difference in its distances to the foci remains constant.
To find the foci, we can use the formula:\[ c = \sqrt{a^2 + b^2} \]
Here, \(c\) represents the distance from the center of the hyperbola to each focus. Adding and subtracting this distance from the center coordinates will give you the actual location of the foci.
The coordinates of the foci help to determine the maximum divergence from the center line and serve as critical points in understanding the path of each hyperbola branch. They are an essential feature that sets hyperbolas apart from other conic sections.
To find the foci, we can use the formula:\[ c = \sqrt{a^2 + b^2} \]
Here, \(c\) represents the distance from the center of the hyperbola to each focus. Adding and subtracting this distance from the center coordinates will give you the actual location of the foci.
The coordinates of the foci help to determine the maximum divergence from the center line and serve as critical points in understanding the path of each hyperbola branch. They are an essential feature that sets hyperbolas apart from other conic sections.
Other exercises in this chapter
Problem 42
Find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$ (y-1)^{2}=-8 x $$
View solution Problem 43
Graph each ellipse and give the location of its foci. $$\frac{x^{2}}{25}+\frac{(y-2)^{2}}{36}=1$$
View solution Problem 43
Convert each equation to standard form by completing the square on \(x\) or \(y .\) Then find the vertex, focus, and directrix of the parabola. Finally, graph t
View solution Problem 44
Graph each ellipse and give the location of its foci. $$\frac{(x-4)^{2}}{4}+\frac{y^{2}}{25}=1$$
View solution