Level 2 - The Focus and Directrix of Parabolas - Connecting Algebra and Geometry Through Coordinates (Geometry)

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Points	Substitution	Simplififcation
$F (0, p)$ and $P (x, y)$	$F P$ $=$ $(x - 0)^{2} + (y - p)^{2}$	$F P$ $=$ $x^{2} + (y - p)^{2}$
$R (x, - p)$ and $P (x, y)$	$R P$ $=$ $(x - x)^{2} + (y - (- p))^{2}$	$R P$ $=$ $(y + p)^{2}$

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Points

Substitution

Simplififcation

F (0, p)

and

P (x, y)

F P

=

(x - 0)^{2} + (y - p)^{2}

F P

=

x^{2} + (y - p)^{2}

R (x, - p)

and

P (x, y)

R P

=

(x - x)^{2} + (y - (- p))^{2}

R P

=

(y + p)^{2}

Focus	Directrix	Equation
$(0, p)$	$y = - p$	$y = \frac{1}{4 p} x^{2}$
$(0, - 1)$	$y = - (- 1)$ $⇕$ $y = 1$	$y = \frac{1}{4 ( - 1 )} x^{2}$ $⇕$ $y = - \frac{1}{4} x^{2}$

Focus

Directrix

Equation

(0, p)

y = - p

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

(0, - 1)

y = - (- 1)

⇕

y = 1

y = \frac{1}{4 ( - 1 )} x^{2}

⇕

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

Transformations of $f (x)$
Horizontal Translations	$Translation right h units, h > 0 y = f (x - h)$
$Translation left h units, h > 0 y = f (x + h)$
Vertical Translations	$Translation up k units, k > 0 y = f (x) + k$
$Translation down k units, k > 0 y = f (x) - k$

Transformations of

f (x)

Horizontal Translations

Translation right h units, h > 0 y = f (x - h)

Translation left h units, h > 0 y = f (x + h)

Vertical Translations

Translation up k units, k > 0 y = f (x) + k

Translation down k units, k > 0 y = f (x) - k

Equation	Focus	Directrix
$y = - \frac{1}{4} x^{2}$	$(0, - 1)$	$y = 1$
$y = - \frac{1}{4} (x - 1)^{2}$	$(1, - 1)$	$y = 1$
$y = - \frac{1}{4} (x - 1)^{2} + 3$	$(1, 2)$	$y = 4$

Equation

Focus

Directrix

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

(0, - 1)

y = 1

y = - \frac{1}{4} (x - 1)^{2}

(1, - 1)

y = 1

y = - \frac{1}{4} (x - 1)^{2} + 3

(1, 2)

y = 4

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Points	Substitution	Simplififcation
$F (- 4, - 2)$ and $P (x, y)$	$F P$ $=$ $(x - (- 4))^{2} + (y - (- 2))^{2}$	$F P$ $=$ $(x + 4)^{2} + (y + 2)^{2}$
$R (2, y)$ and $P (x, y)$	$R P$ $=$ $(x - 2)^{2} + (y - y)^{2}$	$R P$ $=$ $(x - 2)^{2}$

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Points

Substitution

Simplififcation

F (- 4, - 2)

and

P (x, y)

F P

=

(x - (- 4))^{2} + (y - (- 2))^{2}

F P

=

(x + 4)^{2} + (y + 2)^{2}

R (2, y)

and

P (x, y)

R P

=

(x - 2)^{2} + (y - y)^{2}

R P

=

(x - 2)^{2}

Outside of New York, there is a lighthouse on a small island at a distance of $5$ nautical miles from a straight shoreline. One day, Captain Conic has anchored his ship MS Parabola between the lighthouse and the shoreline.

Diagram with the lighthouse marked at the point (0,5) and the ship at (0,3)

When anchored, the distance from the ship to the lighthouse $S L$ is $2$ nautical miles, and the distance from the ship to the shoreline $S S$ is $3$ nautical miles.

When the captain decided to weigh anchor, he gave his helmsman order to sail MS Parabola so that the difference between $S S$ and $S L$ remained constant at $1$ nautical mile. This made the route of the ship form a parabola.

What is the equation of the directrix?

What equation represents the path of the boat?

We will place the lighthouse at (0,5) and we will let the x-axis represent the shoreline.

If the difference between SS and SL is always 1 nautical mile, we can describe the path of the parabola with a focus in (0,5) and a directrix of y=1.

To write the equation that represents the path of the ship, we will consider the standard equation of a vertical parabola with vertex (h,k). VERTICAL PARABOLA y=1/4 p(x- h)^2+ k [0.4em] [-1em] &Vertex: ( h, k) &Focus: ( h, k+ p) &Directrix: y= k- p From the diagram in Part A we know that the vertex is in ( 0, 3). We also know that the directrix is the horizontal line y= 1. Let's add this information to the standard equation. y=1/4 p(x- 0)^2+ 3 [0.4em] [-1em] &Vertex: ( 0, 3) &Focus: ( 0, 3+ p) &Directrix: 1= 3- p To finalize the equation, we have to solve for p.

By substituting p=2 in the expression, we can write the final equation.

Every parabola has something called the latus rectum. This is the line segment that is parallel to the directrix, passes through the focus, and with endpoints on the parabola. Find the length of the latus rectum of the following parabola.

Coordinate plane with the parabola y=1/8*x^2 and its latus rectum

Let's start by finding the equation of the parabola. From the diagram, we see that its vertex is in the origin. It also has a vertical axis of symmetry. This means we can describe the parabola with the following equation. y=1/4 px^2 In this equation, p is the y-coordinate of the focus.

In the diagram we can see that the focus is in the point (0, 2). This means that if we substitute p= 2 we can write the equation of the parabola.

Now we need to find the endpoints of latus rectum. Notice that these points will have the same y-coordinate as the focus.

To determine the x-coordinates, x_A and x_B, we will substitute 2 for y in the equation of the parabola and solve for x.

Notice that Point B lies in the first quadrant. Therefore, this point has the coordinates (4,2). Point A is in the second quadrant which means its coordinates are (- 4,2). The distance between these two points is the difference between the x-coordinates.

The length of latus rectum is 8 units.

Solar energy can be harnessed using mirrors with a parabolic shaped cross section that have a boiler in the focus.

Illustration of a sun rays in a parabolic shaped mirror.

Write an equation on the form

y = a x^{2}

to represent the cross section of a mirror placed

1.8

kilometers from the boiler.

Notice that the equation should be on the form y=ax^2. This means the parabola is vertical and should have its vertex at the origin. The boiler is 1.8 kilometers from the mirror. This means the focus F of the parabola is located at (0, 1.8).

A parabola with its vertex at the origin can be written using the following formula. y=1/4 px^2 The variable p is the y-coordinate of the focus. Therefore, if we substitute p= 1.8 into the equation, we can determine the parabola.

Now we can draw the parabola.

The Focus and Directrix of Parabolas

Catch-Up and Review

Investigating Points that Are Equidistant from a Pair of Points

Parabola

Properties of a Parabola

Hint

Solution

Using the Distance Formula To Calculate Points on a Parabola

Hint

Solution

Equation of a Parabola From Focus and Directrix

Finding the Equation of a Parabola

Answer

Hint

Solution

Investigating Horizontal Parabolas

Answer

Hint

Solution

Parabolas in the Real World

The Focus and Directrix of Parabolas

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