CONIC SECTIONS - THE PARABOLA

Example 1:

Find the coordinates of the focus and the equation of the directrix of the parabola given by

.

First, we will convert the equation of the parabola to standard form.

Since there is a y-term raised to the second power we know that this parabola is of the form .

NOTE: The vertex (h, k) of this parabola is at (0, 0) because there is only a y²-term and no y-term.

Let's multiply both sides of the original equation by 8 to change the coefficient of y² to 1.

 Now, we can write the equation as follows:

The standard form contains a 4p. Currently, 8 sits in its position. Therefore, we can say that

8 = 4p

and p = 2

We can now write the standard form of the original parabolic equation as follows:

where h = 0, k = 0, and p = 2

Coordinates of the Focus (h + p, k)

(0 2, 0) = (2, 0)

Equation of the Directrix x = h p

x = 0 (2)

x = 2

Example 2:

Find the coordinates of the focus and the equation of the directrix of the parabola given by

.

First, we will convert the equation of the parabola to standard form.

Since there is an x-term raised to the second power we know that this parabola is of the form .

NOTE: The vertex (h, k) of this parabola is at (0, 0) because there is only an x²-term and no x-term.

Let's divide both sides of the original equation by 4 to to change the coefficient of x² to 1.

Now, we can write the equation as follows:

The standard form contains a 4p. Currently, sits in its position. Therefore, we can say that

and

We can now write the standard form of the original parabolic equation as follows:

where h = 0, k = 0, and

Coordinates of the Focus (h, k + p)

⁼

Equation of the Directrix y = k p

and

Example 3:

Find the coordinates of the vertex and the focus and the equation of the directrix of the parabola given by

.

First, we will convert the equation of the parabola to standard form.

Since there is an x-term raised to the second power we know that this parabola is of the form .

Let's first multiply both sides of the original equation by 2 to change the coefficient of x² to 1.

Then .

Next, we'll isolate the terms containing x on one side of the equation as follows.

We need (x h)² on the right! We will use the Square Completion Method to alter the right side appropriately.

We will utilize the term 2x. We divide its coefficient 2 by 2, then we raise this quotient to the second power,

That is, .

Now, we add this 1 to BOTH sides of the equation .

1 2y + 1 = x² + 2x + 1

and

Continuing, we get 2 2y = (x + 1)² . 

Now we can begin to rewrite this equation into standard form .

We have (x + 1)², but we can write this as [x (1)]².This is (x h)².

We have 2 2y, but we can write this as 2(y 1).

The standard form contains a 4p. Currently, 2 sits in its position. Therefore, we can say that

2 = 4p

then p =

We can now write the standard form of the original parabolic equation as follows:

[x (1)]² = 4( )(y 1) where h = 1, k = 1, and p =

Coordinates of the Vertex (h, k)

(1, 1)

Coordinates of the Focus (h, k + p)

(1, 1 ) = (1, )

Equation of the Directrix y = k p

y = 1 ( ) = 1 +

and

Example 4:

Find the coordinates of the vertex and focus and the equation of the directrix of the parabola given by

.

First, we will convert the equation of the parabola to standard form.

Since there is a y-term raised to the second power we know that this parabola is of the form .

Let's first multiply both sides of the original equation by 4 to change the coefficient of y² to 1.

Then .

Next, we'll isolate the terms containing y on one side of the equation as follows.

We need (y k)²on the right! We will use the Square Completion Method to alter the right side appropriately.

We will utilize the term 4y. We divide its coefficient 4 by 2, then we raise this quotient to the second power,

That is, .

Now, we add this 4 to BOTH sides of the equation .

4x 16 + 4 = y² 4y + 4

and 4x 12 = y² 4y + 4

Continuing, we get 4x 12 = (y 2)² .

Now we can begin to rewrite this equation into standard form .

We have (y 2)² which is (y k)².

We have 4x 12, but we can write this as 4(x 3).

The standard form contains a 4p. Currently, 4 sits in its position. Therefore, we can say that

4 = 4p

then p = 1

We can now write the standard form of the original parabolic equation as follows:

where h = 3, k = 2, and p = 1

Coordinates of the Focus (h + p, k)

(3 + 1, 2) = (4, 2)

Equation of the Directrix x = h p

x = 3 1

and x = 2