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Chapter 7: Problem 3

Find the vertex, focus, directrix, and axis of the given parabola. Graph the parabola. \(y^{2}=-\frac{4}{3} x\)

Short Answer

Expert verified
Vertex: (0,0). Focus: \((-\frac{1}{3}, 0)\). Directrix: \(x = \frac{1}{3}\). Axis: \(y=0\).

Step by step solution

01

Identify the Parabola Form

The given equation is in the form \(y^2 = -\frac{4}{3}x\). This is the equation of a parabola that opens leftward. The standard form for a horizontal parabola is \((y-k)^2 = 4p(x-h)\). In this equation, \((h, k)\) represents the vertex.
02

Identify the Vertex

For the equation \(y^2 = -\frac{4}{3}x\), we have \(h = 0\) and \(k = 0\) since it can be rewritten as \((y - 0)^2 = -\frac{4}{3}(x - 0)\). Hence, the vertex of the parabola is \((0, 0)\).
03

Determine the Value of p

In the standard form \((y-k)^2 = 4p(x-h)\), \(4p = -\frac{4}{3}\). Therefore, \(p = -\frac{1}{3}\).
04

Identify the Focus

The focus of a horizontal parabola \((y-k)^2 = 4p(x-h)\) is \((h + p, k)\). Plugging in the values, \(h = 0\), \(k = 0\), and \(p = -\frac{1}{3}\), we get \(\left(0 + \left(-\frac{1}{3}\right), 0\right) = \left(-\frac{1}{3}, 0\right)\).
05

Identify the Directrix

The directrix of a horizontal parabola is the line \(x = h - p\). Substituting \(h = 0\) and \(p = -\frac{1}{3}\), the directrix is \(x = 0 - \left(-\frac{1}{3}\right) = \frac{1}{3}\).
06

Identify the Axis of Symmetry

Since the parabola is horizontal as evident from the equation \((y-k)^2 = 4p(x-h)\), the axis of symmetry is a horizontal line that passes through the vertex. Since the vertex is at \((0, 0)\), the axis of symmetry is the line \(y = 0\).
07

Graph the Parabola

To graph the parabola, plot the vertex \((0, 0)\), focus \((-\frac{1}{3}, 0)\), and draw the directrix line \(x = \frac{1}{3}\). Since the parabola opens to the left, sketch the curve extending from the vertex and passing through points symmetric about the focus and directrix.

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Key Concepts

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

Vertex of the Parabola
The vertex of a parabola is a key feature. It is the point where the parabola changes direction. In simpler terms, picture the vertex as the tip of the curve. For the equation \[ y^2 = -\frac{4}{3}x \]we can see it is a horizontal parabola opening leftward. The formula for a horizontal parabola is \[ (y-k)^2 = 4p(x-h) \]where
  • \((h, k)\) is the vertex.
  • Note in our equation, both \(h\) and \(k\) are zero.
Therefore, this parabola's vertex is \((0, 0)\). The vertex acts as a balancing point for the U-shaped curve, and understanding its location helps you graph the parabola more accurately.
Focus of Parabola
The focus of a parabola is an essential part of its structure. Think of the focus as a spotlight shining onto the curve, guiding its shape. To find the focus of a horizontal parabola, use the equation:\[ (y-k)^2 = 4p(x-h) \]Here, the focus is located at the point \((h + p, k)\). For our specific equation \[ y^2 = -\frac{4}{3}x \]we calculated
  • \(p = -\frac{1}{3}\)
Thus, substituting the values, the focus becomes:\[ \left(-\frac{1}{3}, 0 \right) \]This point is crucial because it indicates the direction in which the parabola "opens" or curves. Horizontal parabolas can either open to the left or right, with the focus being pivotal in this determination.
Directrix of a Parabola
The directrix of a parabola is a guideline that affects its shape and orientation. Imagine the directrix as a boundary or invisible line that the parabola cannot pass. For horizontal parabolas of the form \[ (y-k)^2 = 4p(x-h) \]the directrix is a vertical line at\[ x = h - p \]In our case with the parabola \[ y^2 = -\frac{4}{3}x \]we found
  • \(h = 0\),
  • \(p = -\frac{1}{3}\)
So, substituting these values, the directrix becomes\[ x = \frac{1}{3} \]This "line" is significant because the parabola reflects a perfect symmetry around its axis of symmetry, interacting consistently with the directrix. The beauty of parabolas is in this balance between the focus and the directrix.

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