CONIC SECTIONS - THE PARABOLA

Example 1:

Write the geometric standard equation of the following parabola. Then find the coordinates of its vertex.

Since there is a y²- term, we know that this geometric parabola is of the form . It is a parabola either open to the right or open to the left.

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

8 = 4p

and p = 2

We can now write the geometric standard equation of the original parabola as follows:

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

Since p is negative, we can say that the parabola is open to the left.

The coordinates of the vertex are (0, 0).

Example 2:

Write the geometric standard equation of the following parabola. Then find the coordinates of its vertex.

Since there is an x²- term, we know that this geometrixc parabola is of the form . It is a parabola either open up or open down.

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

and

We can now write the geometric standard equation of the original parabola as follows:

where h = 0, k = 0, and

Since p is negative, we can say that the parabola is open down.

The coordinates of the vertex are (0, 0).

Example 3:

Write the geometric standard equation of the following equation of a parabola. Then find the coordinates of its vertex.

Since there is an x²- term, we know that this geometric parabola must be of the form . It is a parabola either open up or open down.

Given an x²- term and an x-term, we have no choice but to use the Square Completion Method. To make this method easier, we will 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.

Here 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

then

Continuing, we factor the right side of the equation to get 2 2y = (x + 1)² . 

Now we can begin to rewrite this equation into the geometric standard equation .

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

We have 2 2y, but we can factor out 2 to get 2(y 1).

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

2 = 4p

then p =

We can now write the geometric standard form of the original parabola as follows:

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

The coordinares of the vertex are ( 1, 1).

Example 4:

Write the geometric standard equation of the following parabola. Then find the coordinates of its vertex.

.

Since there is a y²-term, we know that this geometricd parabola is of the form . It is a parabola either open to the right or open to the left.

Given a y²- term and a y-term, we have no choice but to use the Square Completion Method. To make this method easier, we will 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.

Here we need (y k)² on 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

then 4x 12 = y² 4y + 4

Continuing, we get 4x 12 = (y 2)² .

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

We have (y 2)² which is (y k)².

We have 4x 12, but we can factor out 4 to get 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 geometric standard equation of the original parabola as follows:

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

The coordinares of the vertex are (3, 2).