Problem 7
Question
In Exercises 5–16, find the focus and directrix of the parabola with the given equation. Then graph the parabola. $$ y^{2}=-8 x $$
Step-by-Step Solution
Verified Answer
The focus of the parabola is at \((-2, 0)\) and the equation of the directrix is \(x=2\). The parabola opens to the left, and is symmetrical about the y-axis.
1Step 1: Identify p
The equation \(y^{2}=-8x\) is in the form \(y^{2}=4px\), where \(4p\) is the coefficient of \(x\). Therefore, \(4p = -8\), which implies that \(p = -2\)
2Step 2: Find the focus and directrix
With the value of \(p\) found, the focus and directrix can now be calculated. The focus of a parabola \(y^{2}=4px\) is at \((p, 0)\), so substituting \(p = -2\) gives the focus as \((-2, 0)\). The equation of the directrix of the parabola is \(x=-p\), so substituting \(p = -2\) into this gives \(x=2\) as the equation of the directrix.
3Step 3: Graph the parabola
The parabola can be graphed by plotting a dot at the focus \((-2, 0)\) and drawing a vertical line at \(x=2\) to represent the directrix. Then, draw a smooth curve to represent the parabola, making sure it opens to the left (since \(p<0\)) and is symmetrical about the y-axis.
Key Concepts
Parabola GraphingConic SectionsQuadratic Equations
Parabola Graphing
When graphing parabolas, it's important to understand the role of the focus and directrix. The focus is a fixed point used to define the parabola, while the directrix is a fixed line. For the parabola given by the equation
Graphing starts with plotting the focus and drawing the directrix. You then sketch the curve, making certain it's equidistant from the focus and any point on the directrix at all times. For this example, because (p < 0), the parabola opens to the left. Parabolas can open up, down, left, or right, depending on the sign and position of (p) in the equation. The vertex, at the origin in this case, is the point where the parabola changes direction, and from there, the shape is symmetrical along an axis.
To enhance the graph's accuracy, plot additional points by substituting chosen x-values into the original equation and solving for y. Connect these points in a smooth curve to complete the parabola.
y^2 = -8x, we determined that the focus is (-2, 0) and the directrix is the line (x = 2).Graphing starts with plotting the focus and drawing the directrix. You then sketch the curve, making certain it's equidistant from the focus and any point on the directrix at all times. For this example, because (p < 0), the parabola opens to the left. Parabolas can open up, down, left, or right, depending on the sign and position of (p) in the equation. The vertex, at the origin in this case, is the point where the parabola changes direction, and from there, the shape is symmetrical along an axis.
To enhance the graph's accuracy, plot additional points by substituting chosen x-values into the original equation and solving for y. Connect these points in a smooth curve to complete the parabola.
Conic Sections
Conic sections are curves obtained by intersecting a plane with a double-napped cone. The parabola is one of these conic sections and is characterized by its mirror-symmetry and the fact that it is defined as the set of all points that are equidistant from a single point, the focus, and a line, the directrix.
Other conic sections include the circle, ellipse, and hyperbola, each having unique properties and equations. The parabola differs from the others because it has exactly one focus and directrix, while the other conic sections have two of each. The standard form of a parabola's equation, (y^2 = 4px), directly informs us about its orientation and how it can be graphed. Here, the parameter (p) determines the distance from the vertex to the focus and directrix, as well as the opening's direction.
Other conic sections include the circle, ellipse, and hyperbola, each having unique properties and equations. The parabola differs from the others because it has exactly one focus and directrix, while the other conic sections have two of each. The standard form of a parabola's equation, (y^2 = 4px), directly informs us about its orientation and how it can be graphed. Here, the parameter (p) determines the distance from the vertex to the focus and directrix, as well as the opening's direction.
Quadratic Equations
Quadratic equations are fundamental in algebra and can be written in the form (ax^2 + bx + c = 0), where (a, b, and c) are constants, and (a) is not zero. The graph of a quadratic equation is always a parabola.
The equation (y^2 = -8x) is also a quadratic equation, albeit in a less common form. It can be thought of as an equation with (x) as the dependent variable and (y) squared, indicating the parabola will open horizontally. Quadratic equations are tied to parabolas in that solving a quadratic equation for (y) gives the vertical coordinates for points on the parabola when (x) is known, or vice versa.
Understanding the relationships between a quadratic's coefficients, its graph, and features like the vertex and the axis of symmetry provides crucial intuition for solving and graphing these equations. Techniques such as completing the square or using the quadratic formula provide ways to find the parabola's vertex, focus, and directrix from a quadratic equation.
The equation (y^2 = -8x) is also a quadratic equation, albeit in a less common form. It can be thought of as an equation with (x) as the dependent variable and (y) squared, indicating the parabola will open horizontally. Quadratic equations are tied to parabolas in that solving a quadratic equation for (y) gives the vertical coordinates for points on the parabola when (x) is known, or vice versa.
Understanding the relationships between a quadratic's coefficients, its graph, and features like the vertex and the axis of symmetry provides crucial intuition for solving and graphing these equations. Techniques such as completing the square or using the quadratic formula provide ways to find the parabola's vertex, focus, and directrix from a quadratic equation.
Other exercises in this chapter
Problem 7
Parametric equations and a value for the parameter \(t\) are given. Find the coordinates of the point on the plane curve described by the parametric equations c
View solution Problem 7
Write the appropriate rotation formulas so that in a rotated system the equation has no \(x^{\prime} y^{\prime}\) -term. $$x^{2}+x y+y^{2}-10=$$
View solution Problem 7
find the standard form of the equation of each hyperbola satisfying the given conditions. $$ \text { Foci: }(-4,0),(4,0) ; \text { vertices: }(-3,0),(3,0) $$
View solution Problem 7
In Exercises \(1-18,\) graph each ellipse and locate the foci. $$ \frac{x^{2}}{49}+\frac{y^{2}}{81}=1 $$
View solution