Problem 47
Question
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 \((2+\sqrt{2}, 2)\)
Step-by-Step Solution
Verified Answer
The equation of the parabola is \((y - 2)^2 = 4\sqrt{2}(x - 2)\).
1Step 1: Identify the vertex and standard form
Since the vertex of the parabola is given as \((2, 2)\), the equation of the parabola in vertex form is \((y-k)^2 = 4p(x-h)\), where \((h,k)\) is the vertex of the parabola.
2Step 2: Find the value of p
The directrix is \(x = 2 - \sqrt{2}\) and the focus is \((2 + \sqrt{2}, 2)\). The vertex is exactly halfway between the focus and the directrix. Therefore, the distance from the vertex \((2, 2)\) to the directrix \(x = 2 - \sqrt{2}\) or to the focus \((2 + \sqrt{2}, 2)\) is \(\sqrt{2}\), which is equal to \(|p|\). Thus, \(p = \sqrt{2}\).
3Step 3: Write the complete equation
Using \(p = \sqrt{2}\), and the vertex \((h, k) = (2, 2)\), substitute into the vertex form equation: \((y - 2)^2 = 4\sqrt{2}(x - 2)\).
4Step 4: Express the equation in a standard form if needed
The equation \((y - 2)^2 = 4\sqrt{2}(x - 2)\) might be suitable for some problems, but it can also be expanded if needed to standard quadratic form. However, for vertex equations as described, this step is usually not necessary unless specific reformatting is required.
Key Concepts
Vertex of ParabolaDirectrixFocus of ParabolaVertex Form Equation
Vertex of Parabola
When we talk about the vertex of a parabola, it is helpful to imagine this point as the "center" of the parabola's "U." The vertex is a significant point because it is either the highest or lowest point on the graph, depending on which way the parabola opens. For our specific parabola, the vertex is at the point (2, 2).
- The vertex gives the parabola a specific location on the coordinate plane.
- This is the point \( (h, k) \) in the vertex form equation of a parabola.
Directrix
The concept of the directrix is a bit intriguing but pivotal in defining a parabola. Imagine the directrix as an invisible line that the parabola "wants" to keep equidistant from its opposite side. For our parabola, the directrix is given by the line equation \( x = 2 - \sqrt{2} \).
- The directrix is always perpendicular to the axis of symmetry of the parabola.
- It works along with the focus to help define the shape and direction of the parabola.
Focus of Parabola
The focus is another important feature closely tied with the vertex and directrix. Picture the focus as a point that the parabola is "chasing." For our parabola, the focus is located at \( (2 + \sqrt{2}, 2) \).
- The focus, like the vertex, helps anchor the parabola along its path on a graph.
- The axis of symmetry passes through both the vertex and the focus.
Vertex Form Equation
The vertex form of a parabola equation is a highly efficient way to express a parabola and predicts its behavior. The vertex form is \( (y-k)^2 = 4p(x-h) \), where \( (h, k) \) is the vertex. For our current parabola example, the vertex is \( (2, 2) \) and \( p = \sqrt{2} \).
- Using these, the vertex form equation becomes \( (y-2)^2 = 4\sqrt{2}(x-2) \).
- "\((h, k)\)" pinpoints the vertex precisely, while \(p\) delineates the parabola's openness and direction.
Other exercises in this chapter
Problem 47
For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Directrix: \(x=1 ; e=1\)
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Find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Directrix: \(x=1 ; e=1\)
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For the following exercises, use the given information about the graph of each ellipse to determine its equation. Center at the origin, symmetric with respect t
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For the following exercises, given information about the graph of the hyperbola, find its equation. Vertices at (1,1) and (11,1) and one focus at (12,1)
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