### All Precalculus Resources

## Example Questions

### Example Question #34 : Parabolas

Find the standard form of the equation of a parabola that has its focus at and its directerix at .

**Possible Answers:**

**Correct answer:**

First, we need to figure out if this is a vertical parabola or a horizontal parabola. Since we need to move down units from the focus to the directerix, we know that this is a vertical parabola.

Recall the standard equation of a vertical parabola:

, where is the vertex and is the focal length.

Remember that the distance from the focus to the directerix is . Then, .

With this information, we can then figure out the vertex of the parabola. Since this is a vertical parabola, the -coordinate of the focus and the vertex are the same. To find the -coordinate of the vertex, just move down units from the focus. The vertex is then .

Since the focus is above the vertex, we know that .

Now, from the vertex, we know that and .

Now, plug in all the information to write the standard equation for this parabola:

### Example Question #35 : Parabolas

Find the standard form of the parabola with its vertex at and its focus at

**Possible Answers:**

**Correct answer:**

First, we need to figure out if this is a horizontal or vertical parabola.

Because we need to go down units to get from the vertex to the focus, . We also know at this point that we have a vertical parabola.

Recall the standard equation of a vertical parabola:

, where is the vertex and is the focal length.

Because the question gives us the vertex, we know that and .

Now, plug in all the information to write the standard equation for this parabola:

### Example Question #36 : Parabolas

Find the standard equation of the parabola with its vertex at and its focus at .

**Possible Answers:**

**Correct answer:**

First, we need to determine whether this is a horizontal or a vertical parabola. Because we go left units to get from the vertex to the focus, we know that this is a horizontal parabola with .

Recall the standard equation of a horizontal parabola:

, where is the vertex and is the focal length.

Since the question gives us the vertex, we know that and .

Now, plug all the given information to get the standard equation of a horizontal parabola:

### Example Question #37 : Parabolas

Find the standard form of the equation of the parabola with its focus at and its directerix at .

**Possible Answers:**

**Correct answer:**

First, we need to determine whether this is a horizontal or a vertical parabola. Because we go left units to get from the focus to the directerix, we know that this is a horizontal parabola.

Recall that the distance from the focus to the directerix can be given as . .

Recall the standard equation of a horizontal parabola:

, where is the vertex and is the focal length.

Next, we need to find the vertex. Since this is a horizontal parabola, the -coordinates of the focus and the vertex will be the same. To find the -coordinate, go left units from the focus. The vertex is then located at . Then, and .

Since the focus is located to the right of the vertex, we know that .

Now that we have all the parts needed to write the standard equation for this parabola:

### Example Question #38 : Parabolas

Find the standard form of the equation for the following parabola:

**Possible Answers:**

**Correct answer:**

Recall the standard equation of a horizontal parabola:

, where is the vertex and is the focal length.

Start by isolating the terms.

Complete the square on the left. Make sure to add the same amount to both sides of the equation!

Factor both sides of the equation to get the standard form of a horizontal parabola.

### Example Question #39 : Parabolas

Write the equation for a parabola with a focus at and a directrix at .

**Possible Answers:**

**Correct answer:**

Since the directrix is to the right of the focus, and the vertex should be some distance in between, we know that this parabola opens to the left, and is then negative, in the form:

where p is the distance from the vertex to the focus/directrix, and the vertex is .

Since the directrix is at and the focus has an x-coordinate of 1, the distance from the focus to the directrix is . The vertex is in the middle, or 3 away from each, so . The vertex will be at .

Our equation then is:

or

### Example Question #40 : Parabolas

Write the equation for a parabola with a focus at and a directrix at .

**Possible Answers:**

**Correct answer:**

This equation has a directrix is at , and the y-coordinate of the focus is 1. This means that the distance from the focus to the directrix is 4. The vertex will be halfway between, so 2 away from each. That means that p is 2. If the focus is , then the vertex is .

Since the directrix is a horizontal line, this equation will be in the form , with vertex .

Since the focus is below the directrix, the parabola opens down, and the equation will be negative.

Our equation is: or

### Example Question #41 : Parabolas

Rewrite the following equation for a parabola in standard form:

**Possible Answers:**

**Correct answer:**

To be in standard form, the equation for a parabola must be written in one of the following ways:

OR

THe problem given has the square around the x term, so it's going to end up loking like the standard form on the left.

First, we square the right side

Lastly, we need the y by itself, so we add 3 to both sides

### Example Question #42 : Parabolas

Write the equation for a vertex of that passes through the origin.

**Possible Answers:**

**Correct answer:**

The vertex form for a parabola is given below:

The vertex coordinates are or as given in the problem statement.

Now solve for a using the point it crosses , by plugging the point into the equation.

Plug a back to find the final answer:

### Example Question #43 : Parabolas

Write the equation for a vertex of that passes through the point .

**Possible Answers:**

**Correct answer:**

The vertex form for a parabola is given below:

The vertex coordinates are or as given in the problem statement.

Now solve for a using the point it crosses , by plugging the point into the equation.

Plug a back to find the final answer: