### All Precalculus Resources

## Example Questions

### Example Question #141 : Conic Sections

Find the endpoints of the minor axis of the ellipse with the following equation:

**Possible Answers:**

**Correct answer:**

Recall the standard form of the equation of an ellipse:

, where is the center of the ellipse.

Start by putting the equation in the standard form as shown above.

Group the terms and terms together.

Factor out from the terms and from the terms.

Now, complete the squares. Remember to add the same amount to both sides of the equation!

Subtract from both sides.

Divide by on both sides.

Now factor both terms to get the standard form of the equation of an ellipse.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is horizontal and the endpoints are and .

### Example Question #142 : Conic Sections

Find the endpoints of the minor axis of the ellipse with the following equation:

**Possible Answers:**

**Correct answer:**

Recall the standard form of the equation of an ellipse:

, where is the center of the ellipse.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is vertical and the endpoints are and .

### Example Question #143 : Conic Sections

Find the endpoints of the minor axis of the ellipse with the following equation:

**Possible Answers:**

**Correct answer:**

Recall the standard form of the equation of an ellipse:

, where is the center of the ellipse.

Start by putting the equation in the standard form as shown above.

Group the terms and terms together.

Factor out from the terms and from the terms.

Now, complete the squares. Remember to add the same amount to both sides of the equation!

Add to both sides.

Divide by on both sides.

Now factor both terms to get the standard form of the equation of an ellipse.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is vertical and the endpoints are and .

### Example Question #144 : Conic Sections

Which is **not** the endpoint of a major or minor axis of the ellipse

**Possible Answers:**

**Correct answer:**

The center has an x-coordinate of 4, and the endpoints of the horizontal axis are away from the center. The y-coordinates of these endpoints are the same as the center, -2. So, these points are or and .

The center has a y-coordinate of -2, and the endpoints of the vertical axis are away from the center. The x-coordinates of these endpoints are the same as the center, 4. So, these points are or and .

The only point not listed is , which is , so that's the correct answer choice.

### Example Question #145 : Conic Sections

The center of an ellipse is and the foci are at and . If the length of the minor axis is 22, what are the endpoints of the major axis?

**Possible Answers:**

and

**Correct answer:**

To figure out the endpoints of the major axis, we need to know its length, which we can determine using the equation where a is half the length of the major axis, b is half the length of the minor axis, and c is the distance from the center to the foci.

First, we can determine b. Right now we know that the full length of the minor axis is 22, so half its length is 11. In other words, .

Now we can determine the distance from the foci to the center. The y-coordinates stay the same, so we will be comparing the x-coordinates. The center is and one of the foci is at - they should both be the same distance away, so we can use either for this part. We can do this more informally, too, but we are solving this equation:

adding 4 gives us , so the distance from the center to the foci is 10, or .

Now we can plug this information into the equation:

Since the foci are on the major axis, and since the distance 10 was added to the x-coordinate, we can conclude that the major axis is horizontal.

This means that our endpoints are

### Example Question #146 : Conic Sections

The equation of an ellipse is given by

Find the endpoints of the major and minor axes of the ellipse.

**Possible Answers:**

Major: (4, -8) and (4, 2)

Minor: (1, -3) and (7, -3)

Major: (4, 8) and (4, 2)

Minor: (1, 3) and (7, -3)

Major: (4, 8) and (4, 2)

Minor: (1, -3) and (7, 3)

Major: (-4, -8) and (4, 2)

Minor: (-1, -3) and (7, -3)

Major: (4, -8) and (4, 2)

Minor: (1, 3) and (7, 3)

**Correct answer:**

Major: (4, -8) and (4, 2)

Minor: (1, -3) and (7, -3)

The major and minor axis run through the center of an ellipse in the vertical and horizontal directions. Lets take a look at the equation, which is already in standard form

Because the denominator on the y term is larger, the major axis is in the vertical direction. It's endpoints are located units away from the center, along the verical axis. So, the endpoints of the major axis are given by

and

The minor axis is in the horizontal direction and it's endpoints are located away from the center, along the horizontal axis. So, the endpoints of the minor axis are given by

and

### Example Question #147 : Conic Sections

Express the following equation for a hyperbola in standard form:

**Possible Answers:**

**Correct answer:**

Remember that in order for the equation of a hyperbola to be in standard form, it must be written in one of the following two ways:

Where the point (h,k) gives the center of the hyperbola. In the first option, where the x term is in front of the y term, the hyperbola opens left and right. In the second option, where the y term is in front of the x term, the hyperbola opens up and down. In either case, the distance tells how far above and below or to the left and right of the center the vertices of the hyperbola are. We can see that our equation is of the form given in the second option, as the y term appears first. All we must do to put it in standard form is obtain a 1 on the right side, so we'll divide both sides by 144:

Now we just simplify the fractions on the left side, and the equation of the hyperbola is in standard form:

### Example Question #148 : Conic Sections

Which of the following is an equation for a hyperbola written in standard form?

**Possible Answers:**

**Correct answer:**

In order for the equation of a hyperbola to be in standard form, it must be written in one of the following two ways:

Where the point (h,k) gives the center of the hyperbola, a is half the length of the axis for which it is the denominator, and b is half the length of the axis for which it is the denominator. The easiest requirements of standard form to identify are that the right side of the equation must be 1, there must be subtraction of the terms on the left side and not addition, and the denominators must be different for the x and y terms. Looking at our answer choices, we can see that the following equation is the only one which satisfies these requirements for the standard form of a hyperbola:

### Example Question #151 : Conic Sections

Find the standard equation for a hyperbola defined by the following characteristics:

Vertices: ,

Foci: ,

**Possible Answers:**

**Correct answer:**

Due to the fact that the foci and vertices each share the same x-coordinate, this particle hyperbola can be classified as a vertical transverse axis hyperbola. Thus, it has an equation of the form:

The point (h,k) is the coordinate of the center point, which is the midpoint of the foci. The values a and c are the distances of the midpoint from the vertices and foci, respectively. The b value can be found from a and c via the Pythagorean Theorem.

Putting it all together, the formula for the hyperbola is:

### Example Question #152 : Conic Sections

Write the equation for a hyperbola in standard form given by the equation:

**Possible Answers:**

**Correct answer:**

First, each term should be grouped by like terms (i.e. terms consisting of x's and y's) before factoring. 25 and 16 can be factored out of the x and y groups, respectively, leading to an equation of the form:

Then, one must "complete the squares", or add a term to each group to make them factorable into the form . You should get something that looks like this:

After adding everything together on the right side, condensing the factored out form of the x/y terms, and dividing through by their coefficients, we get the standard form equation:

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