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Chapter 10: Problem 166

For the following exercises, rewrite the given equation in standard form, and then determine the vertex \((V),\) focus \((F)\), and directrix \((d)\) of the parabola. $$y^{2}-6 y+12 x-3=0$$

Short Answer

Expert verified
Vertex: (1, 3), Focus: (-2, 3), Directrix: x = 4

Step by step solution

01

Identify the equation form

The given equation is \(y^2 - 6y + 12x - 3 = 0\). This needs to be rewritten in standard form for a parabola.
02

Isolate the quadratic term

Reorder the equation to focus on the \(y\) terms: \(y^2 - 6y = -12x + 3\).
03

Complete the square for the \(y\) terms

1. Take \(-6y\), identify the coefficient of \(y\) which is \(-6\), divide by \(2\) to get \(-3\), and square it to get \(9\).2. Add and subtract \(9\) on the left side: \(y^2 - 6y + 9 - 9 = -12x + 3\).
04

Simplify the equation

1. The equation becomes: \((y - 3)^2 - 9 = -12x + 3\).2. Rearrange to get: \((y - 3)^2 = -12x + 12\), or \((y - 3)^2 = -12(x - 1)\).
05

Identify conic form parameters

The standard form of a parabola opening sideways is \((y-k)^2 = 4p(x-h)\), where \((h, k)\) is the vertex.1. Comparing \((y - 3)^2 = -12(x - 1)\) with \((y-k)^2 = 4p(x-h)\), we find: \(h = 1\), \(k = 3\), and \(4p = -12\).
06

Calculate parameters and determine vertex

1. Calculate \(p\): Since \(4p = -12\), we have \(p = -3\).2. Vertex \(V\): \((h, k) = (1, 3)\).
07

Calculate focus and directrix

1. Focus \((F)\): The parabola opens leftward (since \(p < 0\)), located at \((h + p, k) = (1 - 3, 3) = (-2, 3)\).2. Directrix \((d)\): The directrix is the vertical line \(x = h - p = 1 + 3 = 4\).

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Key Concepts

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

Vertex
In the context of parabolas, the vertex is a crucial point that serves as both a peak and a symmetrical anchor. It is essentially the point at which the parabola makes its sharpest turn. For the given equation
  • We have derived the vertex, denoted as (V), as the ordered pair (h, k).
  • This is (1, 3) for our equation.
  • The formula (y-k)^2 = 4p(x-h) helps us see that (h, k) = (1, 3).
In graphical terms, the vertex provides the turning point of the parabola. If the parabola opens sideways, as it does in this case, the vertex can be thought of as the maximum or minimum x-value that the parabola will reach.
This vertex also acts as a y-symmetry line, meaning the parabola mirrors itself around the vertical line that intersects the vertex.
Understanding this concept helps in graphing and interpreting parabolas efficiently.
Focus
The focus of a parabola is an internal fixed point and plays a significant role in how the parabola is shaped. In our example:
  • The focus (F) is located at (-2, 3).
  • Determined by the value of 'p' which is -3. <- As p is the distance between the vertex and the focus along the horizontal line.
  • Calculated by adding 'p' to the 'h' value of the vertex (h = 1).
Thus, the focal point is found by evaluating the coordinate (h + p, k).
This equates to (1 - 3, 3) and gives us F = (-2, 3).
The focus of a parabola has the property that any line drawn from a point on the parabola to the focus is equidistant from the directrix, which ensures the unique curve of a parabola.
It serves as a guide to understand the parabola's width and the way branches open.
Directrix
The directrix is a line exterior to the parabola that assists in maintaining its geometric integrity. In coordination with the focus:
  • The directrix is positioned such that every point on the parabola equidistant from the directrix and the focus.
  • With regards to our equation, the directrix is given by the vertical line x = 4.
  • This is calculated using x = h - p, where h = 1 and p = -3, resulting in x = 4.
This particular line seems quite abstract initially but is invaluable in verifying and constructing the parabola's path.
Since parabolas tend to image somewhere between their focus and directrix,
understanding the directrix helps in mastering the graphical aspect of conics.
The elegance of a parabola lies in this simplicity 鈥� maintaining consistent distances between itself, a focus, and a directrix!

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Most popular questions from this chapter

Write the equation of the ellipse in standard form. Then identify the center, vertices, and foci. $$9 x^{2}+y^{2}+54 x-4 y+76=0$$

Graph the hyperbola, noting its center, vertices, and foci. State the equations of the asymptotes. $$y^{2}-x^{2}+4 y-4 x-18=0$$

Write the equation of the parabola in standard form. Then give the vertex, focus, and directrix. $$y^{2}=12 x$$

Graph the equation relative to the \(x^{\prime} y^{\prime}\) system in which the equation has no \(x^{\prime} y^{\prime}\) term. $$6 x^{2}+24 x y-y^{2}-12 x+26 y+11=0$$

For the following exercises, express the equation for two functions, with \(y\) as a function of \(x\) . Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. $$\frac{y^{2}}{9}-\frac{x^{2}}{1}=1$$
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