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Chapter 9: Problem 40

Find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$(y+4)^{2}=12(x+2)$$

Short Answer

Expert verified
The Vertex of the parabola is at \((-2, -4)\), the Focus of the parabola is at \((1, -4)\), and the Directrix of the parabola is at \(x = -5\).

Step by step solution

01

Identify Vertex

The equation of the parabola is \((y+4)^2 = 12(x+2)\). \ Because this is in the form \((y+k)^2 = 4a(x+h)\), we can say that the vertex of the parabola is at the point \((-k, -h)\). \ Substituting the values, the vertex, \(V\) is therefore at the point \((-2,-4)\).
02

Identify Focus

Our value of \(4a\) from the equation is 12. So, \(a = 12/4 = 3\). \ Since our parabola opens to the right (because \(a>0\)), the focus, \(F\), is at the point \((h+a, k)\). \ Substituting, the focus is therefore at the point \((-2+3, -4)\), which simplifies to \((1,-4)\).
03

Identify Directrix

Again, because our parabola opens to the right, the equation of the directrix is of the form \(x = h - a\). \ Substituting from our equation, the directrix, \(D\), is the line \(x = -2 - 3\), which simplifies to \(x = -5\).
04

Graph the Parabola

Begin by plotting the vertex at point \((-2,-4)\). \ The parabola will open to the right, as \(a>0\). \ Next, plot the focus at point \((1,-4)\). The point of the focus is always inside the parabola. \ Plot the directrix as a vertical line at \(x = -5\). The directrix is always outside the parabola. \ Sketch the parabola with the vertex, focus and directrix identified.

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Key Concepts

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

Vertex
The vertex of a parabola is an essential feature that essentially indicates its highest or lowest point, or the point where it changes direction. In the equation \[(y+4)^2 = 12(x+2)\]the expression is structured as \[(y+k)^2 = 4a(x+h)\]. This tells you that the vertex is at the coordinates \(-h, -k\). Substituting the specific values from our equation gives us the vertex at \((-2, -4)\).
The vertex serves as a pivotal point on the graph of the parabola and is crucial for sketching the curve accurately.
  • The vertex will help in understanding the symmetry of the parabola.
  • It is essential for determining other features such as the focus and the directrix.
Recognizing the vertex position makes it easier to depict how the parabola spreads around this central point.
Focus
The focus of a parabola provides valuable information about its orientation and size. In a parabola represented by \((y+k)^2 = 4a(x+h)\), the term \(4a\) helps determine the focus' position relative to the vertex. Let's break it down:
In our example, \(4a=12\) and thus, \(a = 3\). Since the parabola opens to the right, the focus is positioned at \((h+a, k)\). By substituting in the values from the equation, the focus is located at \((1, -4)\).
  • Position of the focus indicates the direction in which the parabola opens.
  • Proximity of the focus to the vertex determines the 'width' of the parabola.
Remember, the focus is always found inside the parabola and helps in shaping the curve.
Directrix
The directrix is a line integral to defining a parabola, serving as a boundary that is always positioned outside it. In our equation, the directrix is determined by the formula \(x = h - a\). For a parabola opening to the right, like ours, this leads to a vertical line.
From our specific equation, upon inserting the values, the directrix is given by \(x = -5\).
  • Every point on a parabola maintains an equal distance from both the focus and its directrix.
  • The directrix helps maintain the locus of the parabola.
Understanding how the directrix works allows us to accurately draw parabolas and demonstrate their symmetry.
Graphing Parabolas
Graphing parabolas involves a few important steps that help visually convey the characteristics of the curve. To graph a parabola, you will need to chart the vertex, focus, and directrix accurately.
Start by plotting the vertex \((-2, -4)\). Next, locate the focus at \((1, -4)\), ensuring it's inside the parabola. Then draw the directrix line at \(x = -5\). With these points marked, sketch the curve such that:
  • The parabola opens to the right.
  • The focus lies within the curve.
  • The directrix remains outside the parabola.
Keeping the critical elements aligned ensures your graph is precise, depicting the parabolic relationship between the given points. This outline will help you visualize the complete parabola on the Cartesian plane.

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Most popular questions from this chapter

In Exercises \(21-40\), eliminate the parameter \(t\). Then use the rectangular equation to sketch the plane curve represented by the given parametric equations. Use arrows to show the orientation of the curve corresponding to increasing values of \(t .\) (If an interval for \(t\) is not specified, assume that \(-\infty < t < \infty .)\) $$x=t, y=-2 t$$

In Exercises \(59-62,\) sketch the plane curve represented by the given parametric equations. Then use interval notation to give each relation's domain and range. $$x=t^{2}-t+6, y=3 t$$

Identify the conic and graph the equation: $$r=\frac{4 \sec \theta}{2 \sec \theta-1}$$

Will help you prepare for the material covered in the next section. a. Make a sketch of an angle \(\theta\) in standard position for which \(\cot 2 \theta=-\frac{7}{24}\) and \(90^{\circ}<2 \theta<180^{\circ}\) b. Use your sketch from part (a) to determine the value of \(\cos 2 \theta\) c. Use the value of \(\cos 2 \theta\) from part (b) and the identities $$\sin \theta=\sqrt{\frac{1-\cos 2 \theta}{2}} \text { and } \cos \theta=\sqrt{\frac{1+\cos 2 \theta}{2}}$$ to determine the values of \(\sin \theta\) and \(\cos \theta\) d. In part (c), why did we not write \(\pm\) before the radical in each formula?

How are the conics described in terms of a fixed point and a fixed line?
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