/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 220 Given a point \(F\) and a line \... [FREE SOLUTION] | 91Ó°ÊÓ

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Given a point \(F\) and a line \(m,\) choose \(m\) as the \(x\) -axis and the line through \(F\) perpendicular to \(m\) as the \(y\) -axis. Let \(F\) have coordinates \((0,2 a)\) (i) Find the equation that defines the locus of points which are equidistant from \(F\) and from \(m\). (ii) Does the equation suggest a more natural choice of axes \(-\) and hence a simpler equation for the locus?

Short Answer

Expert verified
The locus is a parabola \(x^2 = 4a(y - a)\), suggesting the axes shift makes it \(x^2 = 4ay'\).

Step by step solution

01

Understanding the Setting and Choosing Axes

Let the line \(m\) be the \(x\)-axis. Since \(F\) has coordinates \((0, 2a)\), it is located on the \(y\)-axis, 2 units above the \(x\)-axis (\(m\)). We choose axes such that the point \(F\) and the line \(m\) comply with the given coordinates.
02

Identifying the Condition for the Locus

The problem states that the locus of points should be equidistant from \(F\) and the line \(m\). Define an arbitrary point \((x, y)\) as part of the locus. The distance from this point to \(F\) is: \( \sqrt{(x - 0)^2 + (y - 2a)^2} \) and the distance to the line \(m\) is \( |y| \) since it is a horizontal line.
03

Setting Up the Equation for Equidistance

We set the distance from point \((x, y)\) to \(F\) equal to its distance from line \(m\): \[ \sqrt{x^2 + (y - 2a)^2} = |y|. \] To remove the square root, we square both sides of the equation.
04

Squaring the Equation

Squaring both sides, we get:\[ x^2 + (y - 2a)^2 = y^2. \] Expand \((y - 2a)^2\): \[ x^2 + y^2 - 4ay + 4a^2 = y^2. \] Simplifying by canceling \(y^2\) from both sides, we have:\[ x^2 = 4ay - 4a^2 \] or equivalently:\[ x^2 = 4a(y - a). \]
05

Identifying A More Natural Choice of Axes

The equation \( x^2 = 4a(y - a) \) can be recognized as that of a parabola that usually appears in the form \( (y - y_0)^2 = 4p(x - x_0) \), suggesting the axes could be shifted more naturally. If we set the vertex of the parabola at the origin, redefine \(y' = y - a\), then the equation simplifies to \(x^2 = 4ay'\), indicating the vertex form of a parabola.

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Key Concepts

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

Locus
A locus is a set of points that satisfies a particular condition or a group of conditions. In the context of this exercise, we are looking for all the points that are equidistant from a fixed point, known as the focus, and a fixed line, known as the directrix. This specific combination leads to the formation of a parabola, which is a common geometric shape in analytic geometry.
  • The fixed point is labeled as point \( F \) with coordinates \((0, 2a)\).
  • The fixed line \( m \), chosen as the \( x \)-axis, serves as the directrix.
The concept of a locus is fundamental in geometry as it allows for the visualization of geometric shapes and the definition of their properties in the coordinate plane.
Parabola
A parabola is a particular type of curve formed by the locus of points that are equidistant from a fixed point (focus) and a fixed line (directrix). In the exercise, we derived an equation through algebraic manipulation that represents a parabola. The equation we arrived at is: \[ x^2 = 4a(y - a) \]This standard form identifies the vertical parabola with its axis of symmetry parallel to the \( y \)-axis.
The general form of a parabola can be expressed as \[ x^2 = 4py \]where \( p \) represents the distance from the vertex to the focus as well as from the vertex to the directrix.
Understanding parabolas is crucial in coordinate geometry, as they appear frequently in mathematics and related fields, such as physics and engineering, due to their reflective properties and applications.
Coordinate Geometry
Coordinate Geometry, also known as analytic geometry, involves using algebra to analyze geometric problems. It allows the geometric shapes to be translated into algebraic equations, enabling easier manipulation and solution finding.
In this task, by choosing the \(x\)-axis as line \(m\) and maintaining point \(F\) at a specific coordinate, we've engaged in coordinate geometry to comprehend the positioning of the locus and translate distances into algebraic expressions. The result: \[ \sqrt{x^2 + (y - 2a)^2} = |y| \] has illustrated the power of coordinate geometry to simplify and solve complex geometric conditions.
Equidistant
The term equidistant describes a situation where points are an equal distance from a given point or line. In the problem exercise, this principle is central to forming the parabola. Every point on the parabola satisfies the condition of being equidistant from the focus \( F \) and the directrix \( m \). To represent this relationship, we set up the relation: \[ \sqrt{x^2 + (y - 2a)^2} = |y| \] After some algebraic steps, we found it leads us to the familiar parabola equation: \[ x^2 = 4a(y - a) \]. Understanding equidistance in geometry helps in identifying and defining various geometric shapes and their properties.

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Most popular questions from this chapter

Let \(P\) be a point and \(m\) a line not passing through \(P\). Prove that, among all possible line segments \(\underline{P X}\) with \(X\) on the line \(m\), a perpendicular from \(P\) to the line \(m\) is the shortest.

(a)(i) Sketch a unit 2 D-cube as follows. Starting with two unit \(1 \mathrm{D}\) -cubes one directly above the other. Then join up each vertex in the upper 1D-cube to the vertex it corresponds to in the lower 1 D-cube (directly beneath it). (ii) Label each vertex of your sketch with coordinates \((x, y)(x, y=0\) or 1) so that the lower \(2 \mathrm{D}\) -cube has the equation " \(y=0\) " and the upper 2D-cube has the equation " \(y=1 "\). (b)(i) Sketch a unit 3D-cube, starting with two unit \(2 \mathrm{D}\) -cubes \(-\) one directly above the other. Then join up each vertex in the upper \(2 \mathrm{D}\) -cube to the vertex it corresponds to in the lower \(2 \mathrm{D}\) -cube (directly beneath it). (ii) Label each vertex of your sketch with coordinates \((x, y, z)\) (where each \(x, y, z=0\) or 1 ) so that the lower \(2 \mathrm{D}\) -cube has the equation \(" z=0 "\) and the upper \(2 \mathrm{D}\) -cube has the equation \(" z=1 "\). (c)(i) Now sketch a unit \(4 \mathrm{D}\) -cube in the same way \(-\) starting with two unit 3D-cubes, one "directly above" the other. (ii) Label each vertex of your sketch with coordinates \((w, x, y, z)\) (where each \(w, x, y, z=0\) or 1\()\) so that the lower 3 D-cube has the equation \(" z=0\) " and the upper 3 D-cube has the equation " \(z=1\) ".

(a) Find the exact area (in terms of \(\pi\) ) (i) of a semicircle of radius \(r\); (ii) of a quarter circle of radius \(r\) (iii) of a sector of a circle of radius \(r\) that subtends an angle \(\theta\) radians at the centre. (b) Find the area of a sector of a circle of radius \(1,\) whose total perimeter (including the two radii) is exactly half that of the circle itself.

The twelve hour marks for a clock are marked on the circumference of a unit circle to form the vertices of a regular dodecagon ABCDEFGHIJKL. Calculate exactly (i.e. using Pythagoras' Theorem rather than trigonometry) the lengths of all the possible line segments joining two vertices of the dodecagon.

Suppose a regular icosahedron (Problem 189) has edges of length 2. Position vertex \(A\) at the 'North pole', and let \(B C D E F\) be the regular pentagon formed by its five neighbours. (a)(i) Calculate the exact angle between the two faces \(A B C\) and \(A C D\). (ii) How many identical regular icosahedra can one fit together, without overlaps, around a single edge? (b) Let \(\mathcal{C}\) be the circumcircle of \(B C D E F,\) and let \(O\) be the circumcentre of this regular pentagon. (i) Prove that the three edge lengths of the right-angled triangle \(\triangle B O A\) are the edge lengths of the regular hexagon inscribed in the circle \(\mathcal{C}\), of the regular 10 -gon inscribed in the circle \(\mathcal{C},\) and of the regular 5 -gon inscribed in the circle \(\mathcal{C}\). (ii) Calculate the distance separating the plane of the regular pentagon \(B C D E F,\) and the plane of the corresponding regular pentagon joined to the 'South pole'

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