Precalculus - Parabolas

Introduction

• Parabola is the set of all points in a plane equidistant from a fixed point (focus) and a line (directrix).
• Parabola is formed when the intersecting plane to a right circular cone is parallel to the slant height of the cone.
• The eccentricity of a parabola is equal to 1. 
• The axis of symmetry is given by $y=k$ when the parabola opens to the right or opens to the left.
• The axis of symmetry is given by $x=h$ when the parabola opens upward or downward.
• The parabola ${\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{h}\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{4}\mathit{c}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{k}\mathbf{)}$ opens upward when $c>0$ and downward when $c<0.$
• The parabola ${\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{k}\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{4}\mathit{c}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{h}\mathbf{)}$ opens to the right when $c>0$ and opens to the left when $c<0.$
• The general equations of the parabola are as follows:

      ${\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathit{D}\mathit{x}\mathbf{+}\mathit{E}\mathit{y}\mathbf{+}\mathit{F}\mathbf{=}\mathbf{0}$ when the axis of symmetry is parallel to the x-axis.

      ${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{D}\mathit{x}\mathbf{+}\mathit{E}\mathit{y}\mathbf{+}\mathit{F}\mathbf{=}\mathbf{0}$ when the axis of symmetry is parallel to the y-axis. 

• In other references, it is important to note the following general equations of the parabola:

    $\mathit{C}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathit{D}\mathit{x}\mathbf{+}\mathit{E}\mathit{y}\mathbf{+}\mathit{F}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{ }\mathit{C}\mathbf{\ne }\mathbf{0}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{D}\mathbf{\ne }\mathbf{0}$

    $\mathit{A}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{D}\mathit{x}\mathbf{+}\mathit{E}\mathit{y}\mathbf{+}\mathit{F}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{ }\mathit{A}\mathbf{\ne }\mathbf{0}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{E}\mathbf{\ne }\mathbf{0}$

• No matter where we put point C in the parabola, as shown above, the distances between Focus A and point C and between point C and point B are always equal. The line BD is the directrix of the parabola.

Basic Concepts: Standard Equations and Graphs of the Parabola with vertex at (0,0)

The standard equations of the parabola with center at $\left(0,0\right)$ are described below with their corresponding graphs:

1. ${\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{4}\mathit{c}\mathit{x}$, vertex at $\left(0,0\right)$

This is the standard equation of the parabola that opens to the right. 

The graph of $y^2=8x$is shown below.

Directrix is $x=-2$

Focus: $F\left(c,0\right)=A\left(2, 0\right)$

Endpoints of the Latus Rectum:

$L_1\left(c,2c\right)=A\left(2, 4\right)$ and $L_2\left(c,-2c\right)=B\left(2,-4\right)$

 

2. ${\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{4}\mathit{c}\mathit{x}$, vertex at $\left(0,0\right)$

This is the standard equation of the parabola that opens to the left. 

The graph of $y^2=-8x$is shown below.

Directrix is $x=2$

Focus: $F\left(-c,0\right)=C\left(-2, 0\right)$

Endpoints of the Latus Rectum:

$L_1\left(-c,2c\right)=A\left(-2, 4\right)$ and $L_2\left(-c,-2c\right)=B\left(-2,-4\right)$

 

3. ${\mathit{x}}^{\mathbf{2}}\mathbf{=}\mathbf{4}\mathit{c}\mathit{y}$, vertex at $\left(0,0\right)$

This is the standard equation of the parabola that opens upward.

The graph of $x^2=8y$ is shown below.

Directrix is $y=-2$

Focus: $F\left(0, c\right)=C\left(0, 2\right)$

Endpoints of the Latus Rectum:

$L_1\left(2c, c\right)=A\left(4, 2\right)$ and $L_2\left(-2c, c\right)=B\left(-4, 2\right)$

 

4. ${\mathit{x}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{4}\mathit{c}\mathit{y}$, vertex at $\left(0,0\right)$

This is the standard equation of the parabola that opens downward. 

The graph of $x^2=-8y$ is shown below.

Directrix is $y=2$

Focus: $F\left(0,-c\right)=C\left(0,-2\right)$

Endpoints of the Latus Rectum:

$L_1\left(2c,-c\right)=A\left(4,-2\right)$ and $L_2\left(-2c,-c\right)=B\left(-4,-2\right)$

Basic Concepts: Standard Equations and Graphs of the Parabola with vertex at (h,k)

The standard equations of the parabola with vertex at $\left(h,k\right)$ with their corresponding graphs are briefly discussed.

1. ${\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{k}\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{4}\mathit{c}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{h}\mathbf{)}\mathbf{,}$ vertex at $\left(h,k\right)$

This standard equation represents a parabola that opens to the right. 

The graph of ${\left(y-2\right)}^2=8\left(x-3\right)$ is shown below.

Vertex $\left(h, k\right)=\left(3, 2\right)$

Equation of the Directrix: 

$x=h-c\to x=3-2\to x=1$

Focus: $F\left(h+c, k\right)=C\left(5, 2\right)$

Endpoints of the Latus Rectum:

$L_1\left(h+c,k+2c\right)=A\left(5, 6\right)$

$L_2\left(h+c, k-2c\right)=B\left(5,-2\right)$

 

2. ${\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{k}\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{4}\mathit{c}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{h}\mathbf{)}\mathbf{,}\mathbf{ }$vertex at $\left(h,k\right)$

This standard equation represents a parabola that opens to the left.

The graph of ${\left(y-2\right)}^2=-8\left(x-3\right)$ is shown below.

Vertex $\left(h, k\right)=\left(3, 2\right)$

Equation of the DIrectrix:

$x=h-c\to x=3-\left(-2\right)\to x=5$

Focus: $F\left(h-c,k\right)=C\left(1, 2\right)$

Endpoints of the Latus Rectum:

$L_1\left(h-c.k+2c\right)=A\left(1, 6\right)$

$L_2\left(h-c, k-2c\right)=B\left(1,-2\right)$

 

3. ${\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{h}\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{4}\mathit{c}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{k}\mathbf{)}\mathbf{,}\mathbf{ }$vertex at $\left(h,k\right)$

This standard equation represents a parabola that opens upward

The graph of ${\left(x-3\right)}^2=8\left(y-4\right)$ is shown below.

Vertex $\left(h, k\right)=\left(3, 4\right)$

Equation of the Directrix:

$y=k-c\to y=4-c\to y=2$

Focus: $F\left(h, k+c\right)=C\left(3, 6\right)$

Endpoints of the Latus Rectum:

$L_1\left(h+2c,k+c\right)=A\left(7, 6\right)$

$L_2\left(h-2c,k+c\right)=B\left(-1,6\right)$

 

4. ${\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{h}\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{4}\mathit{c}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{k}\mathbf{)}\mathbf{,}\mathbf{ }$vertex at $\left(h,k\right)$

This standard equation represents a parabola that opens downward.

The graph of ${\left(x-3\right)}^2=-8\left(y-4\right)$ is shown below.

Vertex $\left(h, k\right)=\left(3, 4\right)$

Equation of the Directrix:

$y=k-c\to y=4-\left(-2\right)\to y=6$

Focus: $F\left(h, k-c\right)\to C\left(3, 2\right)$

Endpoints of the Latus Rectum:

$L_1\left(h+2c, k-c\right)=A\left(7, 2\right)$

$L_2\left(h-2c, k-c\right)=B\left(-1,2\right)$

Solved Examples

Example 1. Determine the opening, coordinates of the vertex, coordinates of the focus, endpoints, and length of the latus rectum, and equation of the directrix of the parabola ${\mathit{x}}^{\mathbf{2}}\mathbf{=}\mathbf{25}\mathit{y}$.

Solution:

The parabola $x^2=25y$ is in the form of $x^2=4cy$.

The graph of this parabola opens upward since $c>0$.

The vertex is at the origin $\left(0,0\right)$.

The coordinates of the focus follow the form $F\left(0,c\right)$. In this case, we solve for $c$. 

$4c=25\to c=\frac{25}{4} or 6\frac{1}{4}$

Hence, the focus is at $\left(0,6\frac{1}{4}\right)$.

To solve for the endpoints of the latus rectum, we follow the form $L_1\left(2c,c\right) and L_2\left(-2c,c\right)$. 

Hence, the endpoints of the latus rectum are $L_1\left(12\frac{1}{2},0\right) and L_2\left(-12\frac{1}{2},0\right)$.

The length of the latus rectum is given by $L=4c$. 

In this case, we have: $L=4\left(6\frac{1}{4}\right)=25 units$

Since this parabola is in the form of $x^2=4cy$, the equation of the directrix follows the form $y=-c$.

Equation of the Directrix: $y=-6\frac{1}{4}$

 

Example 2. Determine the opening, coordinates of the vertex, coordinates of the focus, endpoints, and length of the latus rectum, and the equation of the directrix of the parabola ${\left(y+3\right)}^2=-9\left(x-7\right)$.

Solution:

The parabola ${\left(y+3\right)}^2=-9\left(x-7\right)$ is in the form of ${\left(y-k\right)}^2=-4c\left(x-h\right)$. 

This parabola opens to the left since $c<0$ and its vertex is at $\left(7,-3\right)$. 

The coordinates of the focus of this parabola follows $F\left(h-c, k\right)$.

To solve for $c$, we have:

$-9=4c\to c=-\frac{9}{4} or -2\frac{1}{4}$

In solving for the coordinates of the focus and the endpoints of the latus rectum, disregard the sign of $c$.

The focus is $\left(4\frac{3}{4},-3\right)$.

The endpoints of the latus rectum are solved using the form $L_1\left(h-c,k+2c\right) and L_2\left(h-c,k-2c\right)$.

In this example, the endpoints of the latus rectum are $L_1\left(4\frac{3}{4},1\frac{1}{2}\right) and L_2\left(4\frac{3}{4},-7\frac{1}{2}\right)$.

The length of the latus rectum is $L=4c=4\left(2\frac{1}{4}\right)=9 units$

The equation of the directrix is $x=h-c\to x=7-\left(-\frac{9}{4}\right)\to x=\frac{37}{4}$

Cheat Sheet

Standard Forms of a Parabola	Coordinates			Axis of Symmetry	Equation of the Directrix
	Vertex	Focus	Latus Rectum
$y^2=4cx$	$\left(0,0\right)$	$\left(c,0\right)$	$\left(c,\pm 2c\right)$	$y=0$	$x=-c$
$x^2=4cy$	$\left(0,0\right)$	$\left(0,c\right)$	$\left(\pm 2c, c\right)$	$x=0$	$y=c$
${\left(y-k\right)}^2=4c\left(x-h\right)$	$\left(h,k\right)$	$\left(h\pm c,k\right)$	$\left(h\pm c,k\pm 2c\right)$	$y=k$	$x=h-c$
${\left(x-h\right)}^2=4c\left(y-k\right)$	$\left(h,k\right)$	$\left(h,k\pm c\right)$	$\left(h\pm 2c, k\pm c\right)$	$x=h$	$y=k-c$

Blunder Areas

• Always inspect the sign of $\mathit{c}\mathbf{,}\mathbf{ }\mathit{h}\mathbf{,}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{k}$ to know the opening of the parabola, the correct coordinates of the vertex, and other properties.
• In solving for the coordinates of the latus rectum and focus, neglect the sign of $c$.
• In solving for the equation of the directrix, always consider the sign of $c$.
• Parabolas in the form of ${\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{\pm }\mathbf{4}\mathit{c}\mathit{x}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }{\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{k}\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{\pm }\mathbf{4}\mathit{c}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{h}\mathbf{)}$ are not representations of quadratic functions. 
• Use the algebraic method in transforming the equation of the parabola in $y^2+Dx+Ey+F=0 or x^2+Dx+Ey+F=0$ in vertex form (standard form). This is done using completing the square.
• The algebraic method and graphical method can both be used to determine, study, and interpret the characteristics/properties of a parabola.