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Chapter 9: Problem 75

Write an equation that is satisfied by the set of points whose distances from the point (-2,0) and the line \(x=2\) are equal.

Short Answer

Expert verified
The equation satisfied by the set of points whose distances from the point (-2, 0) and the line \(x = 2\) are the same is defined by two equations: \(y^2 = -x^2 + 4x\) for \(x >= 2\) and \(x^2 + y^2 = 4x\) for \(x < 2\).

Step by step solution

01

Define the Distances

Let's denote our general point as P(x, y). The distance from point P to the point (-2,0), based on the distance formula, will be \(\sqrt{(x - (-2))^2 + (y - 0)^2}\), which simplifies to \(\sqrt{(x + 2)^2 + y^2}\). The distance from our point P to the line \(x=2\) is the absolute difference of the x-coordinates, which is \(|x - 2|\).
02

Equate the Distances

Since the distances from P to (-2,0) and to the line \(x=2\) are equal, we can set them equal to each other: \(\sqrt{(x + 2)^2 + y^2} = |x - 2|\).
03

Simplify the absolute value

An absolute value can be broken down into two scenarios. Either \(x - 2 >= 0\) (where \(x >= 2\)) or \(x - 2 < 0\) (where \(x < 2\)). Thus our equations are: 1) \(\sqrt{(x + 2)^2 + y^2} = x - 2\) when \(x >= 2\) and 2) \(\sqrt{(x + 2)^2 + y^2} = -x + 2\) when \(x < 2\).
04

Square both sides of the equations

By squaring both sides of the two equations we get rid of the square root and the absolute value: 1) \((x + 2)^2 + y^2 = (x - 2)^2\) when \(x >= 2\) and 2) \((x + 2)^2 + y^2 = (-x + 2)^2\) when \(x < 2\).
05

Simplify the equations

By simplifying those equations will get: 1) \(y^2 = -x^2 + 4x\) when \(x >= 2\) and 2) \( x^2 + y^2 = 4x \) when \(x < 2\). The first equation indicates a downward opening parabola with roots at 0 and 4, while the second equation indicates a circle with radius 2 centered at (2, 0).

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Key Concepts

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

Distance Formula
The distance formula is a fundamental concept in analytic geometry. It helps us determine the distance between two points in a coordinate plane. The formula is derived from the Pythagorean theorem and is written as:- the square root of the sum of squares of the differences in the x-coordinates and y-coordinates.Given two points, \((x_1, y_1)\) and \((x_2, y_2)\), the distance between them, \(d\), is represented by:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This formula allows us to calculate the straight-line distance, or "as-the-crow-flies" distance, between any two points in 2D space. It鈥檚 a powerful tool for solving many problems in geometry.
Parabola Equation
The parabola is a significant shape in analytic geometry characterized by its U-shaped curve. Parabolas can open upwards or downwards and have different equations based on their orientation and position.Here are some key aspects to understand about parabolas:
  • A standard form equation for a vertical parabola is \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants.
  • The value of \(a\) determines the opening direction:
    • If \(a > 0\), the parabola opens upwards, resembling a U.
    • If \(a < 0\), it opens downwards, resembling an inverted U.
  • The vertex of the parabola is its starting "point," determined by the formula \((h, k)\) where \(h = -\frac{b}{2a}\) and \(k\) is the evaluation of the function \(f(h)\).
Understanding the features and equation of a parabola helps in solving various geometric problems where such curves are involved.
Circle Equation
A circle has a unique place in geometry due to its perfectly round shape and symmetry. The standard equation of a circle in a coordinate plane can tell us a lot about its size and position.The fundamental form of a circle's equation is:\[(x - h)^2 + (y - k)^2 = r^2\]where:
  • \((h, k)\) is the center of the circle.
  • \(r\) is the radius.
This equation helps you visualize and graph circles by identifying their center point and how far all points on the circle are from that center point, which is always the radius. It is a critical concept in many fields, letting us find and analyze circular paths or regions.

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Most popular questions from this chapter

In this set of exercises, you will use hyperbolas to study real-world problems. Astronomy The path of a certain comet is known to be hyperbolic, with the sun at one focus. Assume that a space station is located 13 million miles from the sun and at the center of the hyperbola, and that the comet is 5 million miles from the space station at its point of closest approach. Find the equation of the hyperbola if the coordinate system is set up so that the sun lies on the \(x\) -axis and the origin coincides with the center of the hyperbola.

Determine the equations in standard form of two different hyperbolas that satisfy the given conditions. Transverse axis of length \(12 ;\) transverse axis horizontal; one vertex at (6,5)\(;\) slope of one asymptote is -5

Determine the equation in standard form of the hyperbola that satisfies the given conditions. \- Foci at (-3,-6),(-3,-2)\(;\) slope of one asymptote is 1

In this set of exercises, you will use hyperbolas to study real-world problems. Video Games A video game designer is creating a video game about a bank robbcry. In the game, a person robs a bank and then flecs. The robber's path forms one branch of a hyperbola. At the time of the robbery, the bank's security officer is stationed at one focus of the hyperbola and a police cruiser is parked at the opposite vertex (the vertex closer to the opposite focus). If the length of the transverse axis of the hyperbola is 70 millimeters and the foci are 120 millimeters apart, how far from the police cruiser was the bank's security officer when the robbery occurred?

Use a graphing utility to graph the given equation. $$\frac{(x+1)^{2}}{15}-\frac{(y-3)^{2}}{3}=1$$
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