Problem 6
Question
Find the focus and directrix of the parabola with the given equation. Then graph the parabola. $$ y^{2}=4 x $$
Step-by-Step Solution
Verified Answer
The focus of the parabola given by the equation \(y^{2}=4x\) is \((1, 0)\) and the equation of the directrix is \(x=-1\). The parabola opens to the right and is symmetrical about the x-axis.
1Step 1: Identify the coefficient of \(x\)
For the equation of the parabola \(y^{2} = 4px\), the coefficient of \(x\) is 4. Therefore, \(4p = 4\). To find the value of \(p\), divide both sides by 4 so \(p = 1\).
2Step 2: Determine the focus
The focus of the parabola is located at \((p, 0)\). Substituting \(p = 1\) from the first step, we find that the coordinates of the focus are \((1, 0)\).
3Step 3: Determine the directrix
The equation of the directrix is \(x=-p\). Again substituting \(p = 1\) from the first step, we find that the equation of the directrix is \(x=-1\).
4Step 4: Graph the parabola
To graph the parabola, plot its focus and directrix from steps 2 and 3 respectively. Then sketch the parabola so that it opens towards the focus and away from the directrix. The vertex of the parabola is \((0, 0)\). The parabola crosses the y-axis at \((0, 0)\) and the curve is symmetric about the x-axis. The width of the parabola increases with increase in the value of \(|y|\).
Key Concepts
Conic SectionsGraphing ParabolasCoefficient of x in ParabolasSymmetry in Parabolas
Conic Sections
Conic sections are the curves obtained by intersecting a plane with a double napped cone. Depending on the angle at which the plane cuts the cone, the resulting figure can be a circle, an ellipse, a parabola, or a hyperbola. Parabolas are of particular interest in the study of conic sections because they exhibit unique geometric properties and applications, such as their appearance in the paths of projectiles under gravity, in satellite dishes, and in the design of automotive headlights. A parabola is the set of all points that are equidistant from a fixed point, known as the focus, and a straight line called the directrix.
Graphing Parabolas
To graph a parabola, it is essential to identify its vertex, focus, directrix, and axis of symmetry. The vertex is the point where the parabola changes direction, and it's located midway between the focus and the directrix. For the equation of a parabola in standard form, such as \(y^{2}=4px\), one can determine the direction in which the parabola opens—vertical or horizontal—and its width. Here, the graph will be symmetric about the x-axis and will open rightwards since the squared term is \(y\). After plotting the focus and the directrix, the next step is to draw a smooth curve that grows wider as it moves away from the vertex, ensuring it mirrors across the axis of symmetry.
Coefficient of x in Parabolas
The coefficient of \(x\) in the equation of a parabola, represented as \(4p\) in \(y^2 = 4px\), determines the parabola's distance from the vertex to the focus and to the directrix. This coefficient, referred to as the focal parameter or focal width, affects the 'tightness' or 'spread' of the parabola. A larger absolute value of the coefficient results in a wider parabola, while a smaller value leads to a narrower one. In the example given, \(4p = 4\), and solving for \(p\) which yields \(p = 1\), informs us that the parabola is relatively narrow, spreading out slowly from the vertex.
Symmetry in Parabolas
Symmetry in parabolas is a crucial aspect of their graphing and algebraic formulation. The axis of symmetry is a vertical or horizontal line (depending on the orientation of the parabola) that divides the parabola into two mirror-image halves. For the equation \(y^2 = 4px\), the axis of symmetry is the x-axis since the variable \(y\) is squared. This means that for every point on the parabola, there is another point mirroring it across the x-axis. The vertex lies on this line of symmetry, and the parabola's shape is symmetrically reflected across it. The understanding of symmetry helps in accurately sketching parabolas and understanding their reflective properties, which are utilized in numerous optical applications.
Other exercises in this chapter
Problem 5
In Exercises 5-12, find the standard form of the equation of each hyperbola satisfying the given conditions. Foci: \((0,-3),(0,3) ;\) vertices: \((0,-1),(0,1)\)
View solution Problem 6
In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$$
View solution Problem 6
Find the standard form of the equation of each hyperbola satisfying the given conditions. Foci: \((0,-6),(0,6) ;\) vertices: \((0,-2),(0,2)\)
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