Level 1 - 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}

Use the Distance Formula to write the equation of the parabola.

Parabola with a directrix of y = -1 and a focus of (0,1)

An upside down parabola with a directrix of y = 4 and a focus of (0,-4)

Directrix : y = 7 Focus : (0, - 7)

Vertex : Directrix : (0, 0) y = - 9

A parabola can be viewed as every point in a plane that is equidistant from a fixed point called the focus and a fixed line called the directrix. From the diagram, we can identify both the focus and the directrix of the parabola.

Using the Distance Formula, we can write one expression for FP and one for PD. Since these distances are congruent, we can equate these expressions and solve for y.

Distance of FP

By substituting the endpoints of FP we can create an expression for this distance.

Distance of PD

To determine PD, we will substitute D(x,-1) and P(x,y) into the Distance Formula.

Solving for y

As already mentioned, these distances are congruent. Therefore, we can equate them and then solve for y to obtain the equation of the parabola.

Like in Part A, we will first identify the focus and directrix of the given parabola.

Next, we need expressions for FP and PD.

Distance of FP

Distance of PD

Solving for y

As was done in Part A, we will equate the distances and solve for y.

Let's draw a parabola with the given characteristics.

As was done in previous parts, we will now create expressions for FP and PD.

Distance of FP

Distance of PD

Solving for y

Let's equate the distances and solve for y.

Remember that any segment between the focus and an arbitrary point P on the parabola is congruent with the segment from point P to the directrix.

This means the vertex of the parabola must bisect the vertical line segment that we can draw between the directrix and the focus. Therefore, if the directrix is at y=- 9, the focus must be at F(0,9).

Now we can proceed in the same way as in the previous parts.

Distance of FP

Distance of PD

Solving for y

Like in Parts A through C, we will equate the distances and solve for y to obtain the equation of the parabola.

Which of the given characteristics describe a parabola that opens downward? Explain your reasoning.

A : B : C : D : Focus : (0, - 6) Directrix : y = 6 Focus : (0, - 2) Directrix : y = 2 Focus : (0, 6) Directrix : y = - 6 Focus : (0, - 1) Directrix : y = 1

Every parabola that opens downward has a focus that is below the directrix. In the diagram it is possible to graph parabolas with the given characteristics.

We can see that the characteristics A, B, and D result in parabolas that opens downward.

Consider the following parabola with focus at $F (0, - 9)$ and vertex at the origin $V (0, 0) .$

Parabola with focus at F(0,-9), vertex at V(0,0) and an arbitrary point at P(x,y)

Which of the following options are possible coordinates of point

P

that we see in the graph. Explain your reasoning.

A : C : E : (6, - 1) (- 4, - \frac{4}{9}) (- 8, \frac{1 6}{9}) B : (- 3, - \frac{1}{4}) D : (- 1, \frac{1}{3 6}) F : (- 2, - \frac{1}{1 8})

Notice that point P is in the third quadrant of the coordinate plane. This means it has a negative x- and y-coordinate. Therefore we can immediately discard the points A, D, and E. &A: (6,-1) && B: (- 3,- 14 ) [0.4em] &C: (- 4,- 49) && D: (- 1, 136) [0.4em] &E: (-8, 169) && F: (- 2,- 118) Since we have been given the focus and vertex of the parabola, we can find its equation using the following formula. y=1/4px^2 Keep in mind that this formula describes a parabola with its vertex at the origin and with a vertical axis of symmetry. In the formula, p represents the y-coordinate of the focus, which we know is -9. Therefore, by substituting - 9 for p we can obtain the equation of the parabola.

Now we have to substitute each of the remaining points into the equation and see if we end up with a true statement.

Point	Substitute	Evaluate
B( 3, -1/4)	-1/4? =- 1/36( 3)^2	-1/4= -1/4 ✓
C( -4, -4/9)	-4/9? =- 1/36( - 4)^2	-4/9= -4/9 ✓
F( -2, -1/8)	-1/8? =- 1/36( - 2)^2	-1/8= -1/9 *

Now we can strike the remaining option that does not fall on the parabola. &A: (6,-1 ) && B: (- 3,- 14) [0.4em] & C: (- 4,- 49) && D: (- 1, 136) [0.4em] &E: (-8,- 169) && F: (- 2,- 118) The points B and C fall on the parabola.

The distance from point $A$ to the directrix is $2$ units. Write an equation of the parabola.

Parabola that passes through the points A(x,1) and V(0,0)

In the diagram we see that the parabola has its vertex in the origin and has a vertical axis of symmetry. This means we can describe the parabola with the following equation. y=1/4px^2 In this equation, p represents the y-coordinate of the focus. We can also obtain p by finding the opposite number to the directrix. This is because the focus and directrix is equidistant from the vertex. Focus:& (0,p) Directrix:& y=- p Since we know the distance from A to the directrix, we can figure out the equation of the directrix by subtracting 2 from 1. y=1-2 ⇔ y=-1 When we know the equation of the directrix, we can find p by solving the equation y=- p.

Now we can determine the equation of the parabola.

Consider the following parabola.

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

What is its focus

F ?

What is the equation of its directrix?

In the given equation, the variable that has been raised to the power of two is x. y= 1/12x^2 This means the parabola will have a vertical axis of symmetry. We also see that the vertex of the parabola is at the origin. This means the equation of the parabola can be described with the following equation. y= 1/4 px^2 The variable p represents the y-coordinate of the focus. Since the vertex is at the origin, we can write the coordinates of the focus as (0, p). Therefore, to find p, we can equate the coefficient of x in the formula, with the coefficient of x in the given equation. l Given equation → 1/12&= 1/4 p ← Formula Let's solve for p.

The focus is at (0, 3).

From Part A, we know that p=3. However, the opposite number to p also describes the directrix of the parabola. This is because the focus and directrix are equidistant from the vertex. Focus:& (0,3) Directrix:& y=- 3 This means the directrix is y=-3.

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

Distance of FP

Distance of PD

Solving for y

Distance of FP

Distance of PD

Solving for y

Distance of FP

Distance of PD

Solving for y

Distance of FP

Distance of PD

Solving for y

The Focus and Directrix of Parabolas

Recommended exercises

	9 Theory slides
	9 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions