Problem 39
Question
For the following exercises, graph the parabola, labeling the focus and the directrix. \(x^{2}+8 x+4 y+20=0\)
Step-by-Step Solution
Verified Answer
Vertex: (-4, -1), Focus: (-4, -2), Directrix: y = 0.
1Step 1: 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.
2Step 2: 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 \((x + 4)^2 = -4y - 4\).
3Step 3: 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 \(p = -1\).
4Step 4: 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 \(y = 0\).
5Step 5: 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.
Key Concepts
Focus of ParabolaDirectrix of ParabolaCompleting the SquareVertex Form of Parabola
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
- \((x-h)^2 = 4p(y-k)\)
- \((h, k)\)
- \((h, k+p)\)
- \((-4, -2)\)
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:
- \(y = k - p\)
- \((-4, -1)\)
- \(y = 0\)
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 \(x^2 + 8x\)
- The half of 8 is 4, and its square is 16
- \((x + 4)^2 - 16\)
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:
- \((x + 4)^2 = -4(y + 1)\)
- \((x-h)^2 = 4p(y-k)\)
- \((h, k)\) = (-4, -1)
Other exercises in this chapter
Problem 38
For the following exercises, graph the given ellipses, noting center, vertices, and foci. \(\frac{x^{2}}{2}+\frac{(y+1)^{2}}{5}=1\)
View solution Problem 39
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
View solution Problem 39
For the following exercises, graph the given ellipses, noting center, vertices, and foci. \(4 x^{2}-8 x+16 y^{2}-32 y-44=0\)
View solution Problem 40
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
View solution