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Chapter 10: Problem 6

Find the vertex, focus, and directrix of the parabola and sketch its graph. $$ (y-2)^{2}=2 x+1 $$

Short Answer

Expert verified
Vertex: \((-\frac{1}{2}, 2)\); Focus: \((0, 2)\); Directrix: \(x = -1\).

Step by step solution

01

Convert the Equation to Standard Form

The given parabola equation is \((y-2)^2 = 2x+1\). We rewrite it to resemble the standard form of a parabola equation that opens horizontally: \((y-k)^2 = 4p(x-h)\). Start by rewriting the equation as \((y-2)^2 = 2(x+\frac{1}{2})\).
02

Identify Vertex Coordinates

The standard form of a horizontal parabola is \((y-k)^2 = 4p(x-h)\). From the equation \((y-2)^2 = 2(x+\frac{1}{2})\), compare and extract \(h = -\frac{1}{2}\) and \(k = 2\). Thus, the vertex is \((-\frac{1}{2}, 2)\).
03

Find the Value of p

In the standard horizontal form \((y-k)^2 = 4p(x-h)\), \(4p\) must equal the coefficient of \(x\), which is 2. Hence, \(4p = 2\), so \(p = \frac{1}{2}\).
04

Determine Focus Coordinates

For a parabola that opens horizontally, the focus is located \(p\) units from the vertex along the x-axis. Since the parabola opens to the right, the focus is given by \((h+p, k)\). Therefore, the focus is \((-\frac{1}{2} + \frac{1}{2}, 2) = (0, 2)\).
05

Identify the Equation of the Directrix

The directrix of a horizontally opening parabola is a vertical line \(p\) units before the vertex along the x-axis. The equation is \(x = h - p\). For this parabola, \(x = -\frac{1}{2} - \frac{1}{2} = -1\).
06

Sketch the Graph

The vertex is \((-\frac{1}{2}, 2)\), the focus at \((0, 2)\), and the directrix is the line \(x = -1\). The parabola opens to the right, so sketch it with the vertex at \((-\frac{1}{2}, 2)\), tending towards the focus and staying equidistant from it and the directrix.

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Key Concepts

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

Vertex
In the realm of parabolas, the vertex is a crucial point that symbolizes where the parabola changes direction. It's like the tip or the peak, depending on how the parabola opens. For the equation \((y-2)^2 = 2(x+\frac{1}{2})\), which resembles the standard form \((y-k)^2 = 4p(x-h)\), the vertex is identified as \((h, k)\).
  • Extract \(h\) and \(k\) by comparing the equation. Here, \(h = -\frac{1}{2}\) and \(k = 2\).
  • Thus, the vertex is \((-\frac{1}{2}, 2)\).
The vertex gives a starting reference point when sketching or analyzing the parabola. It functions as the midpoint between the focus and the directrix, guiding the direction the parabola opens.
Focus
The focus of a parabola is a magnificent point. It's not on the curve itself but immensely important to its form. This point, along with the directrix, helps define the set of coordinates that make up a parabola.
  • For a horizontally oriented parabola \((y-k)^2 = 4p(x-h)\), the focus is \((h+p, k)\).
  • In our case, since \(h = -\frac{1}{2}\), \(p = \frac{1}{2}\), and \(k = 2\), the focus becomes \((0, 2)\).
The magic behind the focus is that every point on the parabola is equidistant from it and the directrix. This balance is the hallmark of the parabola's symmetrical structure.
Directrix
The directrix may sound complex, but it's simply a straight line. This line, together with the focus, dictates the parabola's shape.
  • For a parabolic equation in the form \((y-k)^2 = 4p(x-h)\), the directrix has the equation \(x = h - p\).
  • Applying these to our specific example, where \(h = -\frac{1}{2}\) and \(p = \frac{1}{2}\), gives us the directrix \(x = -1\).
Imagine the parabola as a bridge. The vertex and focus are critical components, but the directrix provides the unseen support. Every point on the parabola maintains an equal distance to the focus and this imaginary support line, ensuring a perfect and balanced structure.
Parabola equation
The parabola equation encodes all the details needed to sketch or understand this curve's nature. Its standard forms reveal insights into how a parabola opens and its key characteristics:
  • A horizontal opening parabola uses the form \((y-k)^2 = 4p(x-h)\).
  • Similarly, a vertical opening uses \((x-h)^2 = 4p(y-k)\).
Using the equation \((y-2)^2 = 2(x+\frac{1}{2})\), we've unraveled its secrets:
  • It mirrors the horizontal form, indicating it opens sideways, to the right.
  • From this, the elements like \(4p = 2\) and position details \((h, k)\), \(p\) were used to locate the vertex, focus, and directrix.
Understanding this equation requires seeing beyond just the numbers and symbols. Recognize the pattern and symmetry it guarantees, essential to mastering the artwork of parabolas.

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