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Chapter 8: Problem 39

For the following exercises, graph the parabola, labeling the focus and the directrix. \(x^{2}+8 x+4 y+20=0\)

Short Answer

Expert verified
Vertex: (-4, -1), Focus: (-4, -2), Directrix: y = 0.

Step by step solution

01

Convert to Standard Form

The given equation is \(x^2 + 8x + 4y + 20 = 0\). First, isolate the \(y\)-terms by moving them to one side: \(x^2 + 8x + 20 = -4y\). This rearranges to \(x^2 + 8x = -4y - 20\) which will help us when completing the square to convert into standard form.
02

Complete the Square

To complete the square for the \(x\)-terms, take half of the coefficient of \(x\), square it, and add/subtract that inside the equation. Coefficient is 8, half is 4, square is 16: \((x^2 + 8x + 16) - 16 = -4y - 20\). Simplifying: \((x + 4)^2 - 16 = -4y - 20\). Thus, we have \((x + 4)^2 = -4y - 4\).
03

Refactor into Vertex Form

The equation now is in standard form: \((x + 4)^2 = -4(y + 1)\). Here, the vertex form \((x-h)^2 = 4p(y-k)\) indicates the vertex \((h, k)\) is \((-4, -1)\) and \(4p = -4\), so \(p = -1\).
04

Determine Focus and Directrix

For the parabola \((x-h)^2 = 4p(y-k)\), the focus is given by \((h, k + p)\) and the directrix is \(y = k - p\). Thus, focus is \((-4, -2)\) and directrix is \(y = 0\).
05

Sketch the Graph

Using the vertex \((-4, -1)\), focus \((-4, -2)\), and directrix \(y = 0\), plot the parabola. The vertex lies directly halfway between the focus and directrix, indicating that the parabola opens downwards.

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Key Concepts

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

Focus of Parabola
The focus of a parabola is a specific point where all points on the parabola are equidistant from the directrix and follow the reflective property of parabolas. It serves as a guide point from which you can draw the curve of the parabola. In the equation
  • \((x-h)^2 = 4p(y-k)\)
The vertex of the parabola is
  • \((h, k)\)
while the parameter \(p\) dictates the distance from the vertex to the focus. For a vertical parabola, like those described by the equation above, the focus can be found using the formula:
  • \((h, k+p)\)
In our example, the calculations show that \(k+p = -2\), giving us a focus of
  • \((-4, -2)\)
This indicates that the focus lies directly beneath the vertex and guides the "opening" direction of the parabola.
Directrix of Parabola
The directrix of a parabola is a straight line that, alongside the focus, helps define and direct the shape of the parabola. In essence, each point on a parabola is equally distant from both the focus and this directrix line. For a vertical parabola, the directrix can be expressed as the equation:
  • \(y = k - p\)
In our particular parabola example, where the vertex is
  • \((-4, -1)\)
we derived that \(p = -1\), leading to a directrix located at
  • \(y = 0\)
The importance of the directrix lies in its ability to work with the focus to ensure that all points equidistant from the focus and the directrix can form the perfect parabolic arc, ensuring symmetry.
Completing the Square
Completing the square is a powerful algebraic technique for converting quadratic equations into a more revealing form that highlights the essential features of a parabola: its vertex and its axis of symmetry. The process involves transforming a quadratic expression into the square of a binomial:
  • For example, take an equation like \(x^2 + 8x\)
Find half of the coefficient of \(x\), square it, and adjust the equation:
  • The half of 8 is 4, and its square is 16
This yields:
  • \((x + 4)^2 - 16\)
This transformation turns the expression into a recognizable binomial squared form, making it easier to visualize and handle when shifting to vertex form.
Vertex Form of Parabola
The vertex form of a parabola is particularly useful for quickly identifying the vertex of the parabola, which serves as a key reference point. In our equation, we achieved the vertex form through completing the square, resulting in:
  • \((x + 4)^2 = -4(y + 1)\)
The general structure of the vertex form is:
  • \((x-h)^2 = 4p(y-k)\)
From this, we can directly extract the coordinates of the vertex:
  • \((h, k)\) = (-4, -1)
This form is extremely advantageous because it not only shows the location of the vertex, \(h\) and \(k\), but also indicates the direction and width of the parabola's opening through the parameter \(p\). This parameter \(p\) provides crucial information on whether the parabola opens upwards or downwards and how 'stretched' it is.

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Most popular questions from this chapter

For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. \(r=\frac{3}{4-4 \cos \theta}\)

For the following exercises, convert the polar equation of a conic section to a rectangular equation. \(r=\frac{3}{2+5 \quad \cos \theta}\)

For the following exercises, convert the polar equation of a conic section to a rectangular equation. \(r(5+2 \quad \cos \theta)=6\)

For the following exercises, find a new representation of the given equation after rotating through the given angle. \(-2 x^{2}+8 x y+1=0, \theta=45^{\circ}\)

For the following exercises, graph the equation relative to the \(x^{\prime} y^{\prime}\) system in which the equation has no \(x^{\prime} y^{\prime}\) term. \(21 x^{2}+2 \sqrt{3} x y+19 y^{2}-18=0\)
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