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Chapter 10: Problem 27

Determine the vertex, focus, and directrix for each parabola. $$y=\frac{1}{4}(x-3)^{2}$$

Short Answer

Expert verified
The vertex is at (3, 0), the focus is at (3, 1), and the directrix is the line y = -1.

Step by step solution

01

Write the equation in standard form

The given equation is already in vertex form, which is similar to the standard form for a parabola opening upwards or downwards:\[ y = a(x-h)^2 + k \]Here, the equation is \( y = \frac{1}{4}(x-3)^2 \), so we identify \( h = 3 \), \( k = 0 \), and \( a = \frac{1}{4} \).
02

Determine the vertex

The vertex form of the parabola \( y = a(x-h)^2 + k \) gives the vertex directly as \( (h, k) \).Thus, the vertex is at \( (3, 0) \).
03

Find the value of p

The parameter \( p \) is related to \( a \) by the equation \( a = \frac{1}{4p} \). Solving for \( p \) with \( a = \frac{1}{4} \):\[ \frac{1}{4} = \frac{1}{4p} \]\[ p = 1 \]
04

Determine the focus

The focus of a parabola \( y = a(x-h)^2 + k \) is given by \( (h, k+p) \). Here, \( h = 3 \), \( k = 0 \), and \( p = 1 \), so the focus is at:\[ (3, 0+1) = (3, 1) \]
05

Determine the directrix

The directrix of a parabola \( y = a(x-h)^2 + k \) is given by the line \( y = k-p \). With \( k = 0 \) and \( p = 1 \), the directrix is:\[ y = 0-1 = -1 \]

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Key Concepts

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

Understanding the Vertex Form of a Parabola
The vertex form of a parabola is a very useful equation. It allows us to easily identify the vertex of the parabola. A parabola in vertex form is written as: \[ y = a(x-h)^2 + k \]In this equation,
  • \(a\) determines the width and direction of the parabola.
  • \(h\) and \(k\) are the coordinates of the vertex.
For example, consider the equation you provided:\[ y = \frac{1}{4}(x-3)^2 \]Here, by comparing the given equation with the standard vertex form, we can see that:
  • \(h = 3\)
  • \(k = 0\)
  • \(a = \frac{1}{4}\)
Thus, the vertex of this parabola is at (3, 0). This is a significant first step as knowing the vertex gives us crucial information about the shape and position of the parabola.
Determining the Focus of the Parabola
The focus of a parabola is another important feature. It is a point from which distances are measured to define the parabola.To find the focus, we need to determine the parameter \(p\). The relation between \(a\) and \(p\) is:\[ a = \frac{1}{4p} \]Given \(a = \frac{1}{4}\), we solve for \(p\):\[ \frac{1}{4} = \frac{1}{4p} \]\[ p = 1 \]With \(h = 3\), \(k = 0\), and \(p = 1\), we can find the focus using the formula for the coordinates of the focus in the vertex form equation:\[ (h, k + p) \]So, the focus is at (3, 1). This point lies inside the parabola and helps in defining its reflective properties.
Finding the Directrix of the Parabola
The directrix is a line that is used along with the focus to define a parabola. Every point on the parabola is equidistant to the focus and the directrix.For the equation \[ y = a(x-h)^2 + k \]the directrix is given by the line:\[ y = k - p \]Substituting the values we have:\[ k = 0 \]\[ p = 1 \]The directrix can be found as:\[ y = 0 - 1 = -1 \]Thus, the directrix is the horizontal line \( y = -1 \). This line, along with the focus, provides a complete description of the geometric properties of the parabola.

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Most popular questions from this chapter

Find the vertex, axis of symmetry, \(x\) -intercept, \(y\) -intercepts, focus, and directrix for each parabola. Sketch the graph, showing the focus and directrix. $$x=-\frac{1}{2} y^{2}-y-4$$

Find the vertex, axis of symmetry, \(x\) -intercepts, \(y\) -intercept, focus, and directrix for each parabola. Sketch the graph, showing the focus and directrix. $$y=\frac{1}{3} x^{2}-x$$

Use a graphing calculator to solve each problem. The graph of \(x=-y^{2}\) is a parabola opening to the left with vertex at the origin. Find two functions whose graphs will together form this parabola and graph them on your calculator.

Use completing the square to rewrite the equation in one of the standard forms for a conic and identify the conic. $$4 x^{2}+5 y^{2}+2 x-y-1=0$$

Use the discriminant to identify the type of conic without rotating the axes. $$2 x^{2}+3 x y+2 y^{2}+x-y-2=0$$
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