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Chapter 8: Problem 19

For the following exercises, rewrite the given equation in standard form, and then determine the vertex \((V),\) focus \((F),\) and directrix \((d)\) of the parabola. \((y-4)^{2}=2(x+3)\)

Short Answer

Expert verified
Vertex at (-3, 4), focus at (-2.5, 4), directrix is x = -3.5.

Step by step solution

01

Rewrite in Standard Form

The given equation is \((y-4)^2 = 2(x+3)\). Recognize that this equation is in the form \((y-k)^2 = 4p(x-h)\), where \(h = -3\) and \(k = 4\), with \(4p = 2\). Solve for \(p\) by dividing both sides of the equation by 4, resulting in \(p = \frac{1}{2}\). This gives us the standard form as \((y-4)^2 = 2(x+3)\), which is already set accordingly.
02

Identify the Vertex

The vertex \(V\) of the parabola is given by \( (h, k) \). From the standard form \((y-4)^2 = 2(x+3)\), we have \(h = -3\) and \(k = 4\). Therefore, the vertex \(V\) is at \((-3, 4)\).
03

Find the Focus

The focus \(F\) is found using \(h + p = x\)-coordinate and the same \(k\). Since \(p = \frac{1}{2}\) and the parabola opens rightwards, the focus is \((h+p, k) = (-3 + \frac{1}{2}, 4) = (-2.5, 4)\).
04

Determine the Directrix

The directrix is a vertical line \(x = h - p\). Using \(h = -3\) and \(p = \frac{1}{2}\), the directrix is \(x = -3 - \frac{1}{2} = -3.5\).

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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 a critical point that you must understand, as it serves as the "turning point" for the parabola. This is where the direction of the parabola changes.
For a parabola written in standard form as \[(y-k)^2 = 4p(x-h)\], the vertex is located at the point \[(h, k)\].
In our example, the equation is given as \((y-4)^2 = 2(x+3)\). By recognizing this equation type and comparing it against the standard form, the values of \(h\) and \(k\) become apparent:
  • \(h = -3\)
  • \(k = 4\)
Therefore, the vertex is at \((-3, 4)\).
Remember:
  • The vertex is an essential feature for sketching the parabola.
  • It can indicate whether the parabola opens left/right or up/down depending on the standard equation format used.
Focus
The focus of a parabola is a point located inside the curve itself, and it plays a vital role in defining the shape and position of the parabola. Essentially, it's one of the fixed points that the parabola is "focused" upon.
In standard form \((y-k)^2 = 4p(x-h)\), the focus is located at \((h + p, k)\) when the parabola opens left or right.
Given that in our example the expression is \((y-4)^2 = 2(x+3)\) with \(p\) calculated as \(\frac{1}{2}\), we find the focus as follows:
  • \(x\)-coordinate: \(-3 + \frac{1}{2} = -2.5\)
  • \(y\)-coordinate: \(4\)
Hence, the focus is at \((-2.5, 4)\).
Key Points about the Focus:
  • The focus helps in understanding the directrix, another critical line associated with a parabola.
  • Its position relative to the vertex influences the width and position of the parabola.
Directrix
The directrix of a parabola is a line perpendicular to the axis of symmetry and serves as a reference point for constructing and understanding the parabola. It is just as important as the focus, as it helps in determining the "distance property" unique to parabolas.
For a parabola expressed in the equation \((y-k)^2 = 4p(x-h)\), the directrix can be found using the equation: \(x = h - p\).
Using our current example:
  • We have \(h = -3\)
  • We already calculated \(p = \frac{1}{2}\)
The directrix is thus: \(x = -3 - \frac{1}{2} = -3.5\).
Important Notes about the Directrix:
  • The directrix line sits outside the curve of the parabola, contrasting the inner focus.
  • Each point on the parabola is equidistant from the focus and the directrix, forming the defining property of a parabola.
Standard Form
The standard form of a parabola involves rewriting the equation of the parabola so that its properties like the vertex, focus, and directrix can be easily identified. The standard forms for a horizontal or vertical parabola help quickly visualize and solve practical and theoretical problems involving parabolas.
For a horizontal parabola, the standard form is: \((y-k)^2 = 4p(x-h)\).
Here's why it's useful:
  • Easy identification: It allows you to quickly obtain \(h\), \(k\), and \(p\) which are crucial for calculating the vertex, focus, and directrix.
  • Organization: Presents the equation in a uniform way, simplifying computation and interpretation.
In the explained example, the equation \((y-4)^2 = 2(x+3)\) is already in standard form, which means:
  • \(h = -3\)
  • \(k = 4\)
  • Since \(4p = 2\), we find \(p = \frac{1}{2}\)
This simple rearrangement shows the harmony of various parabola properties, making this form integral for easy calculations.

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Most popular questions from this chapter

For the following exercises, determine the angle \(\theta\) that will eliminate the \(x y\) term and write the corresponding equation without the \(x y\) term. \(9 x^{2}-3 \sqrt{3} x y+6 y^{2}+4 y-3=0\)

For the following exercises, convert the polar equation of a conic section to a rectangular equation. \(r=\frac{5}{5-11} \sin \theta\)

For the following exercises, graph the equation relative to the \(x^{\prime} y^{\prime}\) system in which the equation has no \(x^{\prime} y^{\prime}\) term. \(16 x^{2}+24 x y+9 y^{2}-130 x+90 y=0\)

For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. \(r(4-5 \sin \theta)=1\)

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