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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
Graph the parabola with vertex at (-4,-1), focus at (-4,-2), and directrix at y=0.

Step by step solution

01

Write the equation in standard form

The given equation is \( x^2 + 8x + 4y + 20 = 0 \). First, let's rewrite it in the standard form of a parabola, which is \( (x-h)^2 = 4p(y-k) \). Begin by isolating the terms involving \( x \). We do this by subtracting 20 from both sides and then subtracting \( 4y \) from both sides, leading to: \[ x^2 + 8x = -4y - 20 \] .
02

Complete the square

For the expression \( x^2 + 8x \), complete the square. Take half of the linear coefficient of \( x \), which is 4, and square it: \( 4^2 = 16 \). Add and subtract 16 inside the equation: \[ x^2 + 8x + 16 = -4y - 20 + 16 \] This simplifies to: \[ (x + 4)^2 = -4y - 4 \].
03

Solve for y

Simplify the equation you have from the previous step: \[ (x + 4)^2 = -4(y + 1) \]Thus, the equation of the parabola is in vertex form \( (x - h)^2 = 4p(y - k) \) where \( h = -4 \), \( k = -1 \) and \( 4p = -4 \). Therefore, \( p = -1 \).
04

Identify the vertex, focus, and directrix

In the vertex form \( (x - h)^2 = 4p(y - k) \), our parabola's vertex is \( (h, k) = (-4, -1) \). The focus is \((h, k + p) = (-4, -2)\) because \( p = -1 \). The directrix is the line \( y = k - p \), which is \( y = 0 \).
05

Graph the parabola, focus, and directrix

To graph the parabola, plot the vertex at \((-4, -1)\) and sketch the parabola opening downwards because \( p = -1 \) (a negative value). Position the focus at \((-4, -2)\), just below the vertex and draw the directrix as the horizontal line \( y = 0 \), above the vertex. Ensure the parabola is symmetric with respect to the vertex.

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Key Concepts

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

Vertex Form
The vertex form of a parabola helps us understand the shape and position of the parabola on a graph. It's a way of expressing the equation that highlights the vertex of the parabola, which is a critical point. The vertex form is given by the equation:
\[ (x - h)^2 = 4p(y - k) \]
In this equation:
  • \( (h, k) \) represents the vertex of the parabola. It's where the parabola changes direction.
  • \( p \) is a parameter that affects the "width" and "direction" of the parabola's opening.
  • The sign of \( p \) tells us which way the parabola opens: positive for up, negative for down.
By converting our equation to this form, we can easily identify the vertex and direction of the parabola. In our example, the vertex is \((-4, -1)\) and since \( p = -1 \), the parabola opens downward.
Focus and Directrix
Every parabola has an associated point called the "focus" and a line called the "directrix". These geometric attributes help define the parabola in precise mathematical terms.
  • The focus is a point situated along the axis of symmetry of the parabola. For the parabola \((x - h)^2 = 4p(y - k)\), the focus is located at \((h, k + p)\).
  • The directrix is a line perpendicular to the axis of symmetry and is placed "opposite" of the focus relative to the vertex. Its equation is \(y = k - p\).
For our given equation, the parabola鈥檚 focus is at \((-4, -2)\) and the directrix is \(y = 0\). These elements are used for graphing or deeper geometric analysis of the parabola.
Completing the Square
Completing the square is a method used to transform a quadratic equation into a more useful form, typically the vertex form for parabolas. This technique simplifies finding the vertex and rewriting the equation.
To complete the square:
  • Start with a quadratic expression such as \(x^2 + bx\).
  • Take half of the coefficient of \(x\), then square it. For instance, if the coefficient is 8, half is 4, and its square is 16.
  • Add and subtract this square within the equation to maintain balance.
With our parabola, we complete the square on \(x^2 + 8x\) by adding and subtracting 16, transforming the equation to \((x + 4)^2 = -4(y + 1)\), which is in vertex form. This allows us to easily identify the vertex, making graphing the parabola easier.

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Most popular questions from this chapter

For the following exercises, find the equation of the parabola given information about its graph. Vertex is (2,2)\(;\) directrix is \(x=2-\sqrt{2},\) focus is \((2+\sqrt{2}, 2)\)

For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation. $$ V(4,-3), \text { Endpoints }\left(5,-\frac{7}{2}\right),\left(3,-\frac{7}{2}\right) $$

For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Directrix: \(x=-\frac{1}{4} ; e=\frac{7}{2}\)

Recall from Rotation of Axes that equations of conics with an \(x y\) term have rotated graphs. For the following exercises, express each equation in polar form with \(r\) as a function of \(\theta\). $$ x y=2 $$

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. $$ r=\frac{3}{1+2 \cos \theta} $$
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