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Use the process of completing the square to transform the given equation to a standard form. Then name the corresponding curve and sketch its graph. $$ 3 x^{2}-10 y^{2}+36 x-20 y+68=0 $$

Short Answer

Expert verified
The equation represents a hyperbola with center \((-6, -1)\).

Step by step solution

01

Rearrange the Equation

Reorder the terms in the given equation to group the x and y terms together. We have the equation:\[3 x^2 + 36 x - 10 y^2 - 20 y + 68 = 0\]Group the x terms and the y terms separately:\[(3 x^2 + 36 x) + (-10 y^2 - 20 y) + 68 = 0\]
02

Factor and Complete the Square for x

Factor out the coefficient of \(x^2\) from the x terms:\[3(x^2 + 12x) - 10(y^2 + 2y) + 68 = 0\]To complete the square for the expression \(x^2 + 12x\), take half of the coefficient of x (which is 12), square it (\((12/2)^2 = 36\)), and add and subtract it inside the grouping:\[3(x^2 + 12x + 36 - 36) - 10(y^2 + 2y) + 68 = 0\]Rewriting yields:\[3((x + 6)^2 - 36) - 10(y^2 + 2y) + 68 = 0\]Simplify:\[3(x + 6)^2 - 108 - 10(y^2 + 2y) + 68 = 0\]
03

Factor and Complete the Square for y

Continue by completing the square for the y terms inside the equation. Factor out the coefficient of \(y^2\):\[3(x + 6)^2 - 108 - 10(y^2 + 2y + 1 - 1) + 68 = 0\]For the expression \(y^2 + 2y\), find half of the coefficient of y, square it (\((2/2)^2 = 1\)), and add and subtract it:\[3(x + 6)^2 - 108 - 10((y + 1)^2 - 1) + 68 = 0\]Simplify the expression:\[3(x + 6)^2 - 10(y + 1)^2 = 108 - 68\]
04

Solve for Standard Form

Combine and simplify the last few terms:\[3(x + 6)^2 - 10(y + 1)^2 = 40\]Divide each term by 40 to express the equation as:\[\frac{3}{40}(x + 6)^2 - \frac{10}{40}(y + 1)^2 = 1\]Simplify the fractions:\[\frac{3}{40}(x + 6)^2 - \frac{1}{4}(y + 1)^2 = 1\]To present the standard form, divide the entire equation bythe coefficient of the leading term:\[\frac{(x + 6)^2}{ rac{40}{3}} - \frac{(y + 1)^2}{4} = 1\]This aligns with the standard form of a hyperbola:\[\frac{(x + h)^2}{a^2} - \frac{(y + k)^2}{b^2} = 1\]where \((h, k) = (-6, -1)\), and distinguishing the equation as a hyperbola.
05

Identify and Sketch the Curve

The equation has been converted into the standard form of a hyperbola. The center of the hyperbola is given by \((-6, -1)\). The hyperbola opens horizontally, as indicated by the standard form. Sketch the hyperbola with its center at \((-6, -1)\), and asymptotes determined from the slopes \(\pm\frac{b}{a}\).

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Key Concepts

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

Hyperbola
A hyperbola is a type of conic section formed by intersecting a plane with a double-cone in such a way that the resulting shape curves outward, forming two separate branches. An essential feature of a hyperbola is its two foci, located symmetrically on either side of its center.
To visualize a hyperbola, imagine two identical mirror-image curves that extend indefinitely in opposite directions. They resemble two open arcs that diverge from each other. The standard form of a hyperbola equation is:
  • Horizontal Hyperbola: \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\)
  • Vertical Hyperbola: \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\)
Here, \((h, k)\) represents the center of the hyperbola, \(a\) is the distance from the center to the vertices along the axis of symmetry, and \(b\) is the distance from the center to the co-vertices. The hyperbola's asymptotes, invisible lines that the branches approach but never meet, indicate the overall direction and degree of curvature.
Conic Sections
Conic sections are the different shapes formed by the intersection of a plane with a cone. These shapes include circles, ellipses, parabolas, and hyperbolas. Each shape has its own unique equation and properties. The type of conic is determined by the angle at which the intersecting plane cuts through the cone.
A clear understanding of conic sections is vital in geometry and various fields like physics and engineering. Let's explore the characteristics of each conic section:
  • Circle: A set of points equidistant from a center point.
  • Ellipse: Like a stretched circle, where the sum of the distances from two foci to any point on the ellipse remains constant.
  • Parabola: Formed when a plane is parallel to an edge of the cone, with a directrix and a focus determining its shape.
  • Hyperbola: As discussed earlier, consists of two symmetric open curves.
Understanding conic sections helps in modeling real-world phenomena, such as planetary orbits in astronomy or parabolic antennas in telecommunications.
Equation Transformation
Equation transformation involves modifying the form of an equation to reveal more about its properties. One common transformation technique is completing the square, often used to convert quadratic equations into standard forms, which makes it easier to identify their key features.
In completing the square, the quadratic equation is manipulated so that it represents a perfect square trinomial. For example, to complete the square for an expression like \(x^2 + bx\):
  • Take half of the coefficient of \(x\) (which is \(b/2\)).
  • Square it.
  • Add and subtract this square to convert the expression into a perfect square form \((x + b/2)^2 - (b/2)^2\).
This process is vital when deriving the standard forms of conic sections, like transforming equations to identify a hyperbola's center, vertices, and asymptotes. Practicing equation transformation, such as completing the square, enhances your ability to solve a variety of mathematical problems by simplifying expressions or equations.

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Most popular questions from this chapter

Consider a bridge deck weighing \(\delta\) pounds per linear foot and supported by a cable, which is assumed to be of negligible weight compared to the bridge deck. The cable section \(O P\) from the lowest point (the origin) to a general point \(P(x, y)\) is shown in Figure \(13 .\) The forces acting on this section of cable are \(H=\) horizontal tension pulling at \(O\) \(T=\) tangential tension pulling at \(P\) \(W=\delta x=\) weight of \(x\) feet of bridge deck For equilibrium, the horizontal and vertical components of \(T\) must balance \(H\) and \(W,\) respectively. Thus, $$ \frac{T \sin \phi}{T \cos \phi}=\tan \phi=\frac{\delta x}{H} $$ That is, $$ \frac{d y}{d x}=\frac{\delta x}{H}, \quad y(0)=0 $$ Solve this differential equation to show that the cable hangs in the shape of a parabola. (Compare this result with that for the unloaded hanging cable of Problem 53 of Section \(6.9 .)\)

An asteroid from deep space is sighted from the earth moving on a parabolic path with the earth at the focus. When the line from the earth to the asteroid first makes an angle of \(90^{\circ}\) with the axis of the parabola, the asteroid is measured to be 40 million miles away. How close will the asteroid come to the earth (see Problem 33 )? Treat the earth as a point.

Name the conic (horizontal ellipse, vertical hyperbola, and so on ) corresponding to the given equation. \(\frac{-x^{2}}{9}+\frac{y}{4}=0\)

Find the equations of the tangent and the normal lines to the given parabola at the given point. Sketch the parabola, the tangent line, and the normal line. \(x^{2}=4 y,(4,4)\)

Find the coordinates of the focus and the equation of the directrix for each parabola. Make a sketch showing the parabola, its focus, and its directrix. \(y^{2}=-12 x\)

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