If a parabola has the focus at ( 0 , − 1 ) \left(0,-1\right) ( 0 , − 1 ) and a directrix line y = 1 y=1 y = 1 , find the standard equation for the parabola.

− 4 y = x 2 -4y=x^2 − 4 y = x 2

4 ( y − 1 ) = x 2 4\left(y-1\right)=x^2 4 ( y − 1 ) = x 2

− 4 ( y + 1 ) = x 2 -4\left(y+1\right)=x^2 − 4 ( y + 1 ) = x 2

If a parabola has the focus at ( 2 , 4 ) \left(2,4\right) ( 2 , 4 ) and a directrix line x = − 4 x=-4 x = − 4 , find the standard equation for the parabola.

12 ( x + 1 ) = ( y − 4 ) 2 12\left(x+1\right)=\left(y-4\right)^2 12 ( x + 1 ) = ( y − 4 ) 2

− ( x + 1 ) = ( y − 4 ) 2 -\left(x+1\right)=\left(y-4\right)^2 − ( x + 1 ) = ( y − 4 ) 2

4 ( x − 1 ) = ( y + 4 ) 2 4\left(x-1\right)=\left(y+4\right)^2 4 ( x − 1 ) = ( y + 4 ) 2

8. Conic Sections

Parabolas

8. Conic Sections

Parabolas: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Parabolas can be derived from slicing a cone with a tilted plane, resulting in a focus point and a directrix line, both equidistant from the vertex. The standard form of a parabola is expressed as $y^{2} = 4 p x$ for horizontal parabolas and $x^{2} = 4 p y$ for vertical ones. The sign of p determines the direction of the opening, with positive values indicating upward or rightward openings, while negative values indicate downward or leftward openings.

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Parabolas as Conic Sections

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Parabolas as Conic Sections Video Summary

In the study of conic sections, the parabola is a unique shape that can be visualized by slicing a three-dimensional cone with a tilted plane. Understanding parabolas involves identifying two key components: the focus and the directrix. The focus is a specific point on the graph, while the directrix is a line. Both the focus and the directrix are equidistant from the vertex of the parabola, although they may be oriented in different directions.

When analyzing a parabola, the orientation (opening upwards or downwards) determines the location of the focus and directrix. For a parabola that opens upwards, the focus is positioned above the vertex, while the directrix is below it. Conversely, if the parabola opens downwards, the focus is below the vertex, and the directrix is above it.

The standard form of a parabola's equation is crucial for finding the focus and directrix. The general equation can be expressed as:

$ y = \frac{1}{4p}(x - h)^2 + k $

In this equation, $ (h, k) $ represents the vertex, and $ p $ is a value that indicates the distance from the vertex to the focus and the directrix. If the vertex is at the origin, the equation simplifies to:

$ y = \frac{1}{4p}x^2 $

To find the value of $ p $, one can compare the coefficient of $ y $ in the equation to $ 4p $. For example, if $ 4p = 4 $, then $ p = 1 $. This means that for a parabola with a vertex at the origin, the focus would be located 1 unit above the vertex, and the directrix would be 1 unit below it.

When given a specific parabola equation, the first step is to identify the vertex. For instance, if the equation is $ y - 2 = 2(x - 1)^2 $, the vertex can be determined as $ (1, 2) $. Next, the value of $ p $ can be calculated by recognizing that $ 4p = 8 $, leading to $ p = 2 $. This indicates that the focus is located 2 units above the vertex at the point $ (1, 4) $.

To determine the width of the parabola, one can move left and right from the focus by $ 2p $ units. In this case, moving 4 units from the focus gives points at $ (-3, 4) $ and $ (5, 4) $. Connecting these points with a smooth curve forms the parabola.

Finally, the directrix is found by moving down $ p $ units from the vertex. Since $ p = 2 $, the directrix is located at $ y = 0 $. This process illustrates how to effectively find the focus, directrix, and graph the parabola, enhancing understanding of this conic section.

Problem

Graph the parabola $-4\left(y+1\right)=\left(x+1\right)^2$ , and find the focus point and directrix line.

Problem

If a parabola has the focus at $\left(0,-1\right)$ and a directrix line $y=1$ , find the standard equation for the parabola.

$4y=x^2$

$4\left(y-1\right)=x^2$

$-4y=x^2$

$-4\left(y+1\right)=x^2$

example

Parabolas as Conic Sections Example 1

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Parabolas as Conic Sections Example 1 Video Summary

To analyze parabolas, we utilize the standard equation of a parabola, which can be expressed as $4p(y - k) = (x - h)^2$. In this equation, $(h, k)$ represents the vertex of the parabola, while $p$ indicates the distance from the vertex to the focus and the directrix. If the equation lacks $h$ and $k$, the parabola is centered at the origin, simplifying the equation to $4py = x^2$.

For example, consider a parabola represented by $4py = x^2$. Here, since there are no $h$ and $k$ values, the vertex is at the origin $(0, 0)$. Setting $4p = 16$ allows us to solve for $p$, yielding $p = 4$. A positive $p$ indicates that the parabola opens upward. Consequently, the focus is located at $(0, 4)$ and the directrix is the line $y = -4$.

In another case, if we have a parabola defined by $4py = \frac{1}{3}$, we again identify the vertex at the origin. Setting $4p = \frac{1}{3}$ leads to $p = \frac{1}{12}$. The positive value of $p$ confirms that the parabola opens upward, placing the focus at $(0, \frac{1}{12})$ and the directrix at $y = -\frac{1}{12}$.

When the equation includes $h$ and $k$, such as $4p(y - 1) = (x - 2)^2$, the vertex shifts to $(2, 1)$. Here, setting $4p = 8$ gives $p = 2$. The focus, therefore, is at $(2, 3)$ (moving up 2 units from the vertex), while the directrix is the line $y = -1$ (2 units below the vertex).

For a parabola like $-12y = (x + 1)^2$, we first rewrite it to identify the vertex. The vertex is at $(-1, 0)$ since $h = -1$ and $k = 0$. Setting $4p = -12$ results in $p = -3$, indicating that the parabola opens downward. Thus, the focus is at $(-1, -3)$ and the directrix is the line $y = 3$ (3 units above the vertex).

In summary, the vertex, focus, and directrix of a parabola can be determined using the standard form of the equation. The value of $p$ not only indicates the direction in which the parabola opens but also helps in locating the focus and directrix relative to the vertex. Understanding these relationships is crucial for solving problems involving parabolas effectively.

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Horizontal Parabolas

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Horizontal Parabolas Video Summary

In the study of parabolas, we typically encounter vertical parabolas that open either upwards or downwards. However, when we consider horizontal parabolas, the equations and characteristics change slightly, primarily involving the switching of the x and y variables. This transformation leads to a new understanding of how these parabolas behave and how to graph them effectively.

For horizontal parabolas, the general equations are derived from their vertical counterparts. The standard form for a vertical parabola is given by $4py = x^2$, while for a horizontal parabola, it becomes $4px = y^2$. This switch indicates that the orientation of the parabola has changed, with the axis of symmetry now being horizontal. Consequently, the directrix, which is a line that helps define the parabola, is vertical for horizontal parabolas and horizontal for vertical ones.

Understanding the parameter $p$ is crucial, as it determines the direction in which the parabola opens. For vertical parabolas, a positive $p$ indicates an upward opening, while a negative $p$ indicates a downward opening. Conversely, for horizontal parabolas, a positive $p$ means the parabola opens to the right, and a negative $p$ means it opens to the left. This relationship can be visualized on a graph, where positive values typically extend upward and to the right, while negative values extend downward and to the left.

To graph a horizontal parabola, one must follow a systematic approach. First, identify the vertex, which is the center of the parabola. If the equation lacks specific $h$ and $k$ values, the vertex is at the origin (0,0). Next, calculate the value of $p$ from the equation. For example, if the equation is $y^2 = 8x$, we can set $4p = 8$ to find $p = 2$.

Once $p$ is determined, locate the focus, which lies $p$ units in the direction the parabola opens. For a parabola opening to the right, the focus would be at (2,0) if starting from the origin. The width of the parabola can be established by moving $2p$ units up and down from the focus, resulting in points that help define the shape of the parabola.

Finally, the directrix is found by moving $p$ units in the opposite direction of the opening. For a parabola that opens to the right, the directrix would be the line $x = -2$. By connecting these points with a smooth curve, one can accurately represent the horizontal parabola on a graph.

In summary, understanding the differences between vertical and horizontal parabolas, particularly in terms of their equations, orientations, and properties, is essential for mastering their graphical representations and applications in various mathematical contexts.

Problem

Graph the parabola $8\left(x+1\right)=\left(y-2\right)^2$ , and find the focus point and directrix line.

option a

option b

optiuon c

option d

Problem

If a parabola has the focus at $\left(2,4\right)$ and a directrix line $x=-4$ , find the standard equation for the parabola.

$12\left(x+1\right)=\left(y-4\right)^2$

$-\left(x+1\right)=\left(y-4\right)^2$

$12x=y^2$

$4\left(x-1\right)=\left(y+4\right)^2$

example

Horizontal Parabolas Example 1

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Horizontal Parabolas Example 1 Video Summary

In studying parabolas, it's essential to understand their geometric properties, particularly the vertex, focus, and directrix. A horizontal parabola can be represented by the equation:

$ 4p(x - h) = (y - k)^2 $

Here, $ (h, k) $ denotes the vertex, and $ p $ represents the distance from the vertex to the focus and the directrix. A positive $ p $ indicates that the parabola opens to the right, while a negative $ p $ indicates it opens to the left.

For example, consider a horizontal parabola where the equation is given without any terms subtracted from $ x $ or $ y $. This indicates that the vertex is at the origin $ (0, 0) $. If we set $ 4p = 4 $, solving for $ p $ gives $ p = 1 $. Thus, the focus is located at $ (1, 0) $ and the directrix is the line $ x = -1 $.

In another case, if the equation shows a subtraction from $ y $, such as $ y - 2 $, the vertex shifts to $ (0, 2) $. Setting $ 4p = 9 $ results in $ p = \frac{9}{4} $, indicating the focus is at $ \left(\frac{9}{4}, 2\right) $ and the directrix is at $ x = -\frac{9}{4} $.

When the equation includes terms like $ y + 2 $, it’s crucial to recognize that this translates to $ y - (-2) $, thus $ k = -2 $. If $ 4p = 16 $, then $ p = 4 $, placing the focus at $ (8, -2) $ and the directrix at $ x = 0 $.

Lastly, if the equation indicates a subtraction from $ x $ and no subtraction from $ y $, such as $ x - 1 $, the vertex is at $ (1, 0) $. If $ 4p = -2 $, solving gives $ p = -\frac{1}{2} $, meaning the parabola opens to the left. The focus would then be at $ \left(\frac{1}{2}, 0\right) $ and the directrix at $ x = \frac{3}{2} $.

Understanding these principles allows for the effective identification of the vertex, focus, and directrix of horizontal parabolas, which is fundamental in the study of conic sections.

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We have more practice problems on Parabolas

Here’s what students ask on this topic:

What is the standard form equation of a parabola and how is it used to find the focus and directrix?

The standard form equation of a parabola depends on its orientation. For a vertical parabola with the vertex at the origin, the equation is:

y = \frac{1}{4 p} x^{2}

Here, $p$ represents the distance from the vertex to the focus or directrix. For horizontal parabolas, the equation switches $x$ and $y$ :

x = \frac{1}{4 p} y^{2}

To find the focus, move $p$ units in the direction the parabola opens. For the directrix, move $p$ units in the opposite direction. Positive $p$ values indicate upward or rightward opening, while negative values indicate downward or leftward opening.

How do you graph a parabola step by step?

To graph a parabola, follow these steps:

Find the vertex: Use the $h$ and $k$ values from the equation $y = \frac{1}{4 p} x^{2} + k$ .
Calculate $p$ : Solve for $p$ using $4 p$ from the equation.
Locate the focus: Move $p$ units in the direction the parabola opens.
Determine the width: Go $2 p$ units left and right (or up and down for horizontal parabolas) from the focus.
Draw the curve: Connect the points smoothly to form the parabola.
Find the directrix: Move $p$ units opposite the parabola's opening direction.

What is the difference between vertical and horizontal parabolas?

Vertical and horizontal parabolas differ in orientation and equation structure:

Vertical parabolas open up or down, with the equation:
$y = \frac{1}{4 p} x^{2}$
Horizontal parabolas open left or right, with the equation:
$x = \frac{1}{4 p} y^{2}$
The axis of symmetry is vertical for vertical parabolas and horizontal for horizontal parabolas.
The directrix is horizontal for vertical parabolas and vertical for horizontal parabolas.
Positive $p$ values indicate upward or rightward opening, while negative values indicate downward or leftward opening.

How does the value of $p$ affect the shape and orientation of a parabola?

The value of $p$ determines the parabola's orientation and width:

Positive $p$ values indicate upward-opening (vertical parabola) or rightward-opening (horizontal parabola).
Negative $p$ values indicate downward-opening (vertical parabola) or leftward-opening (horizontal parabola).
The absolute value of $p$ affects the distance between the vertex and the focus/directrix, influencing the parabola's width. Larger $p$ values result in wider parabolas, while smaller values create narrower shapes.

What are the steps to find the directrix of a parabola?

To find the directrix of a parabola:

Identify the vertex from the equation. For $y = \frac{1}{4 p} x^{2}$ , the vertex is $(h, k)$ .
Calculate $p$ by solving $4 p$ from the equation.
Determine the direction the parabola opens (up, down, left, or right).
Move $p$ units opposite the opening direction from the vertex. For vertical parabolas, adjust the $y$ -coordinate; for horizontal parabolas, adjust the $x$ -coordinate.
Write the directrix as a line equation: $y =$ or $x =$ .

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Parabolas: Videos & Practice Problems

Parabolas as Conic Sections

Parabolas as Conic Sections Video Summary

Graph the parabola −4(y+1)=(x+1)2-4\left(y+1\right)=\left(x+1\right)^2−4(y+1)=(x+1)2, and find the focus point and directrix line.

If a parabola has the focus at (0,−1)\left(0,-1\right)(0,−1) and a directrix line y=1y=1y=1, find the standard equation for the parabola.

Parabolas as Conic Sections Example 1

Parabolas as Conic Sections Example 1 Video Summary

Horizontal Parabolas

Horizontal Parabolas Video Summary

Graph the parabola 8(x+1)=(y−2)28\left(x+1\right)=\left(y-2\right)^28(x+1)=(y−2)2 , and find the focus point and directrix line.

If a parabola has the focus at (2,4)\left(2,4\right)(2,4) and a directrix line x=−4x=-4x=−4 , find the standard equation for the parabola.

Horizontal Parabolas Example 1

Horizontal Parabolas Example 1 Video Summary

Do you want more practice?

Here’s what students ask on this topic:

What is the standard form equation of a parabola and how is it used to find the focus and directrix?

How do you graph a parabola step by step?

What is the difference between vertical and horizontal parabolas?

How does the value of p affect the shape and orientation of a parabola?

What are the steps to find the directrix of a parabola?

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Graph the parabola $-4\left(y+1\right)=\left(x+1\right)^2$ , and find the focus point and directrix line.

If a parabola has the focus at $\left(0,-1\right)$ and a directrix line $y=1$ , find the standard equation for the parabola.

Graph the parabola $8\left(x+1\right)=\left(y-2\right)^2$ , and find the focus point and directrix line.

If a parabola has the focus at $\left(2,4\right)$ and a directrix line $x=-4$ , find the standard equation for the parabola.

How does the value of $p$ affect the shape and orientation of a parabola?