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Chapter 11: Problem 48

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

Short Answer

Expert verified
Vertex: (2, 3), Focus: (2, 4), Directrix: y = 2

Step by step solution

01

- Identify the Standard Form

The given equation \((x-2)^{2}=4(y-3)\) is in the standard form of a parabola \((x-h)^2 = 4p(y-k)\).
02

- Identify the Vertex

From the standard form, the vertex (\(h, k)\) can be determined directly. Here, \(h = 2\) and \(k = 3\). Therefore, the vertex of the parabola is at \( (2, 3) \).
03

- Determine the Value of p

By comparing \( 4p \) with the coefficient in the equation, \(4 = 4p\), solve for \(p\). Thus, \(p = 1\).
04

- Find the Focus

The focus of a parabola \((x-h)^2 = 4p(y-k)\) is \((h, k + p)\). Substituting the values, we get the focus \( (2, 3+1) = (2, 4) \).
05

- Find the Directrix

The directrix of a parabola \((x-h)^2 = 4p(y-k)\) is given by the line \( y = k - p \). Plugging in the values, we get the directrix: \( y = 3 - 1 = 2 \).
06

- Graph the Parabola

To graph the parabola, plot the vertex \( (2, 3) \), the focus \((2, 4) \), and the directrix \(( y = 2 )\). Sketch the parabola opening upwards with these points in mind.

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Key Concepts

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

Understanding the Parabola Standard Form
The standard form of a parabola is a simplified way to understand its equation and key features. For a parabola that opens upwards or downwards, the standard form is \((x-h)^2 = 4p(y-k)\). Here:
  • \(h\) and \(k\): the coordinates of the vertex
  • \(p\): the distance from the vertex to the focus, which also determines how 'wide' or 'narrow' the parabola is
Identifying the standard form from the equation helps you quickly pinpoint the vertex, focus, and directrix. It simplifies the entire process, making it easier to graph and understand the parabola's orientation.
Vertex Calculation
Finding the vertex of a parabola is straightforward when it's in standard form. The vertex \((h, k)\) represents the point where the parabola changes direction. From the equation \((x-2)^{2}=4(y-3)\), we can identify \h\ and \k\ directly:
  • \h = 2\
  • \k = 3\
Thus, the vertex is \(2, 3\). The vertex is crucial because it acts as a 'central' point from which we measure other key features like the focus and directrix.
Focus Calculation
The focus of a parabola is a point that helps define its shape. In the standard form \((x-h)^2 = 4p(y-k)\), the focus lies at \((h, k + p)\).To find the focus, first determine \p\. In our case, equate \4p\ with the coefficient of the equation:
  • \4p = 4\
  • \p = 1\
Now, use the values for \h\, \k\, and \p\:
  • \h = 2\
  • \k = 3\
  • \p = 1\
Therefore, the focus is \((2, 3+1) = (2, 4)\).
Directrix Calculation
The directrix of a parabola is a line that is perpendicular to its axis of symmetry. In the standard form \((x-h)^2 = 4p(y-k)\), the directrix is given by \y = k - p\.Here鈥檚 how to find it:
  • Identify \k\: \k = 3\
  • Identify \p\: \p = 1\
Plug in the values: \y = 3 - 1\This gives the directrix as \y = 2\.The directrix helps in understanding the geometric properties of the parabola, balancing out with the focus.

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Most popular questions from this chapter

Use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit. The mean distance of Earth from the Sun is 93 million miles. If the aphelion of Earth is 94.5 million miles, what is the perihelion? Find an equation for the orbit of Earth around the Sun.

Use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit. The mean distance of Mars from the Sun is 142 million miles. If the perihelion of Mars is 128.5 million miles, what is the aphelion? Find an equation for the orbit of Mars about the Sun.

Transform the polar equation \(r=6 \sin \theta\) to an equatiol in rectangular coordinates. Then identify and graph the equation.

Hyperbolic Mirrors Hyperbolas have interesting reflective properties that make them useful for lenses and mirrors. For example, if a ray of light strikes a convex hyperbolic mirror on a line that would (theoretically) pass through its rear focus, it is reflected through the front focus. This property, and that of the parabola, were used to develop the Cassegrain telescope in \(1672 .\) The focus of the parabolic mirror and the rear focus of the hyperbolic mirror are the same point. The rays are collected by the parabolic mirror, then reflected toward the (common) focus, and thus are reflected by the hyperbolic mirror through the opening to its front focus, where the eyepiece is located. If the equation of the hyperbola is \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\) and the focal length (distance from the vertex to the focus) of the parabola is \(6,\) find the equation of the parabola.

Find an equation for the hyperbola described. Graph the equation. Vertices at (0,-6) and (0,6)\(;\) asymptote the line \(y=2 x\)
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