Chapter 22: Problem 25
Write each general equation in standard form by completing the square. Find the center, lengths of the semiaxes, foci, slope of the asymptotes, and vertices. $$16 x^{2}-64 x-9 y^{2}-54 y=161$$
Short Answer
Expert verified
The standard form is \(\frac{(x - 2)^2}{9} - \frac{(y + 3)^2}{16} = 1\). Center: (2, -3), Semiaxes: a = 3, b = 4, Foci: (7, -3) and (-3, -3), Slope of asymptotes: \(\pm \frac{4}{3}\), Vertices: (5, -3) and (-1, -3).
Step by step solution
01
Group x and y terms and move constant to the other side
First, we isolate the x-terms and y-terms and move the constant term to the opposite side of the equation. The equation becomes: \[16x^2 - 64x - 9y^2 - 54y = 161\] Rewriting gives: \[16x^2 - 64x - 9y^2 - 54y - 161 = 0\] Now move the constant to the right side: \[16x^2 - 64x - 9y^2 - 54y = 161\]
02
Factor the coefficients of the squared terms
In preparation for completing the square, factor out the coefficient of the squared terms from each binomial: \[16(x^2 - 4x) - 9(y^2 + 6y) = 161\]
03
Complete the square for x and y
To complete the square, add \((\frac{b}{2})^2\) for the x and y terms inside the parentheses (where b is the coefficient of the x or y term). Do this for both binomials (x and y terms): \[16(x^2 - 4x + 4) - 9(y^2 + 6y + 9) - 16(4) + 9(9) = 161\]
04
Simplify the equation
Simplify the equation by performing the operations inside the parentheses, including multiplying the added constants by the factored coefficients: \[16(x - 2)^2 - 9(y + 3)^2 = 161 + 16(4) - 9(9)\] \[16(x - 2)^2 - 9(y + 3)^2 = 161 + 64 - 81\] \[16(x - 2)^2 - 9(y + 3)^2 = 144\]
05
Write in standard form for a hyperbola
To write the equation in standard form for a hyperbola, divide every term by 144: \[\frac{16(x - 2)^2}{144} - \frac{9(y + 3)^2}{144} = 1\] Simplify the fractions: \[\frac{(x - 2)^2}{9} - \frac{(y + 3)^2}{16} = 1\]
06
Find the center, semiaxes, foci, slope of asymptotes, and vertices
The standard form is now compared to the standard form of a hyperbola \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\):- Center: (h, k) = (2, -3)- Lengths of semiaxes: a = 3 (horizontal semiaxis), b = 4 (vertical semiaxis)- Foci: c = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5; coordinates (h \pm c, k) = (2 \pm 5, -3)- Slope of asymptotes for a horizontal hyperbola: \pm b/a = \pm 4/3- Vertices: Vertices are located at \pm a units from the center along the transverse axis; coordinates (h \pm a, k) = (2 \pm 3, -3)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a method used in algebra to transform a quadratic equation into a perfect square trinomial, thereby simplifying the equation and making it easier to solve or analyze. This method is particularly useful when dealing with the equations of conic sections like hyperbolas.
For a given quadratic expression in the form of ax^2 + bx + c, the goal is to turn it into the form of a(x - h)^2 + k, where (h, k) represents the vertex of the parabola. The process involves three main steps:
In our exercise, completing the square enabled us to transform the given equation into a standard form equation by creating perfect square trinomials for x and y, which then could be easily compared to the standard form of a hyperbola.
For a given quadratic expression in the form of ax^2 + bx + c, the goal is to turn it into the form of a(x - h)^2 + k, where (h, k) represents the vertex of the parabola. The process involves three main steps:
- Factor out the coefficient a from the x^2 and x terms, if a ≠ 1.
- Add and subtract the square of half the coefficient of x, which is \(\left(\frac{b}{2a}\right)^2\), inside the brackets.
- Simplify and adjust the equation to reflect the addition and subtraction by incorporating these into the constant term c.
In our exercise, completing the square enabled us to transform the given equation into a standard form equation by creating perfect square trinomials for x and y, which then could be easily compared to the standard form of a hyperbola.
Hyperbola Characteristics
A hyperbola, one of the conic sections, has distinct characteristics that set it apart from other conic shapes like circles, ellipses, and parabolas. Here are some key features of a hyperbola:
In the context of our exercise, identifying these characteristics is crucial for graphing the hyperbola and understanding its geometric properties. We derived the center, lengths of the semiaxes, location of the foci, slope of asymptotes, and coordinates of the vertices by converting the equation into standard form.
- Two Separate Branches: A hyperbola consists of two symmetrical open curves that extend indefinitely.
- Center: The center point of a hyperbola is the midpoint of the line segment connecting its foci and lies at the point of intersection of its axes.
- Axes: A hyperbola has two principal axes - the transverse axis, which passes through the foci, and the conjugate axis, which is perpendicular to the transverse axis.
- Vertices: The points where the hyperbola intersects the transverse axis are called vertices, and they are closest to the center.
- Asymptotes: These are straight lines that the hyperbola approaches but never touches, indicating the direction in which the branches extend.
- Foci: Hyperbolas have two foci, and the distance between each focus and any point on the hyperbola remains constant.
In the context of our exercise, identifying these characteristics is crucial for graphing the hyperbola and understanding its geometric properties. We derived the center, lengths of the semiaxes, location of the foci, slope of asymptotes, and coordinates of the vertices by converting the equation into standard form.
Conic Sections
Conic sections are the curves obtained by intersecting a plane with a double-napped cone at various angles. These shapes include circles, ellipses, parabolas, and hyperbolas, each having unique equations and characteristics. The type of conic is determined by the angle the plane makes with the cone's axis.
Here's how the intersection defines each conic:
Understanding conic sections is key to mastering a wide array of mathematical and scientific problems, especially within coordinate geometry and analytical geometry. In our solved problem, the standard form equation relates to the hyperbola, a specific type of conic section characterized by its subtractive equation format and unique set of features, including two diverging branches and a set of real asymptotes.
Here's how the intersection defines each conic:
- Circle: The plane cuts the cone parallel to the base.
- Ellipse: The plane cuts the cone at an angle steeper than the side of the cone but not parallel to the base.
- Parabola: The plane is parallel to the side of the cone.
- Hyperbola: The plane cuts through both halves of the cone, and the angle is sharper than for an ellipse.
Understanding conic sections is key to mastering a wide array of mathematical and scientific problems, especially within coordinate geometry and analytical geometry. In our solved problem, the standard form equation relates to the hyperbola, a specific type of conic section characterized by its subtractive equation format and unique set of features, including two diverging branches and a set of real asymptotes.