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Chapter 8: Problem 29

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. $$ 3 y^{2}-4 x-6 y+23=0 $$

Short Answer

Expert verified
The vertex is \((5, 1)\), the focus is \((\frac{83}{16}, 1)\), and the directrix is \(x = \frac{77}{16}\).

Step by step solution

01

Rewrite the Equation in Standard Form

Given the equation: \( 3y^2 - 4x - 6y + 23 = 0 \), we need to rewrite it in the standard form for a parabola. First, move terms involving \( x \) to one side:\[3y^2 - 6y = 4x - 23\]Next, complete the square for the \( y \) terms. Start with \( 3(y^2 - 2y) \). To complete the square, take the coefficient of \( y \) in the bracket, divide by 2, square it, and add and subtract inside the parenthesis.The term to add and subtract is \( (\frac{2}{2})^2 = 1 \). With completion, the equation becomes:\[3[(y - 1)^2 - 1] = 4x - 23\]Distribute the \( 3 \):\[3(y - 1)^2 - 3 = 4x - 23\]Add \( 3 \) to both sides:\[3(y - 1)^2 = 4x - 20\]Divide everything by \( 4 \) to isolate \( x \):\[x = \frac{3}{4}(y - 1)^2 + 5\]This equation is now in the form of a parabola opening horizontally: \( x = a(y - k)^2 + h \).
02

Identify the Vertex

From the standard form \( x = \frac{3}{4}(y - 1)^2 + 5 \), the vertex \((h, k)\) can be identified as \((5, 1)\).
03

Determine the Orientation and Coefficient

The coefficient \( a = \frac{3}{4} \) indicates the parabola opens to the right as it is positive. The form \((y-k)^2 = 4px\) equates with \( x = \frac{3}{4}(y - 1)^2 + 5 \), giving \( 4p = \frac{3}{4} \).
04

Calculate the Focus and Directrix

Given \( 4p = \frac{3}{4} \), solve for \( p \):\[p = \frac{3}{16}\]The focus \( F(h+p, k) \) is therefore \((5 + \frac{3}{16}, 1) = (\frac{83}{16}, 1)\).The directrix is the line \( x = h - p = 5 - \frac{3}{16} = \frac{77}{16} \).

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Key Concepts

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

Standard Form of Parabola
The standard form of a parabola provides valuable information about its shape and orientation.
When dealing with a horizontally oriented parabola, like the one we have in this exercise, the standard form is typically written as:
  • \(x = a(y-k)^2 + h\)
In this form:
  • \(a\) determines the direction it opens (right if \(a > 0\), left if \(a < 0\)).
  • \((h, k)\) represents the vertex of the parabola, giving us the precise point where it turns.
To transform a given equation into the standard form, isolate and complete the square for the variable squared (in this case, \(y\)). This process results in a neat expression that reveals essential elements like the vertex, focus, and directrix, all crucial to understanding the parabola's geometric properties.
Vertex of Parabola
The vertex of a parabola is an essential feature as it represents the turning point.
In the context of our parabola described by the standard form \(x = \frac{3}{4}(y - 1)^2 + 5\), the vertex can be directly read as:
  • \((h, k) = (5, 1)\)
Here:
  • \(h\) represents the x-coordinate of the vertex.
  • \(k\) is the y-coordinate.
For horizontally oriented parabolas, the vertex provides a central anchor point from which the parabola stretches outward. Once the equation is in standard form, identifying the vertex becomes straightforward, allowing further calculations for other features such as the focus and directrix.
Focus of Parabola
The focus of a parabola is a unique point that produces the characteristic reflective properties of a parabola.
For our horizontally oriented parabola described by \(x = \frac{3}{4}(y - 1)^2 + 5\), the focus can be calculated using the vertex and the value of \(p\).
  • The formula for the focus in a horizontal parabola is \( (h+p, k) \).
Here, \(4p = \frac{3}{4}\), which means:
  • \(p = \frac{3}{16}\)
Combining this with the vertex, the coordinates of the focus are:
  • \((5 + \frac{3}{16}, 1) = (\frac{83}{16}, 1)\)
The focus is one component of the parabola's vertical cross-section symmetry, making it a magnet for the lines drawn from points on the parabola to this specific focus point.
Directrix of Parabola
The directrix of a parabola is a fixed straight line used to define the conic section's structure.
In the equation \(x = \frac{3}{4}(y - 1)^2 + 5\) with its horizontal opening, the directrix provides balance to the parabola's structure.
  • The directrix is found using the formula \(x = h - p\).
Given \(h = 5\) and \(p = \frac{3}{16}\), the directrix's equation becomes:
  • \(x = 5 - \frac{3}{16} = \frac{77}{16}\)
The directrix lies on the opposite side of the vertex from the focus, maintaining a consistent distance of \(2p\) from it. This geometric balance ensures that the parabola's reflective symmetry is preserved, defining its shape precisely.

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Most popular questions from this chapter

Graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. $$ r(3-2 \sin \theta)=6 $$

An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center.

For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. $$ r=\frac{5}{2+\cos \theta} $$

Graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. $$ r(6-4 \cos \theta)=5 $$

For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation. $$ V(4,-3), \text { Endpoints }\left(5,-\frac{7}{2}\right),\left(3,-\frac{7}{2}\right) $$
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