This equation should be thought of as .

This means that the hyperbola will be determined by a box with x-intercepts at and y-intercepts at .

The hyperbola was incorrectly drawn with the intercepts at instead.

This equation should be thought of as .

This means that the hyperbola will be determined by a box with x-intercepts at and y-intercepts at .

The hyperbola was incorrectly drawn with the intercepts at instead.

The equation for the eccentricity of an ellipse is , where is eccentricity, is the distance from the foci to the center, and is the square root of the larger of our two denominators.

Our denominators are and , so .

To find , we must use the equation , where is the square root of the smaller of our two denominators.

This gives us , so .

Therefore, we can see that

.

The equation for the eccentricity of an ellipse is , where is eccentricity, is the distance from the foci to the center, and is the square root of the larger of our two denominators.

Our denominators are and , so .

To find , we must use the equation , where is the square root of the smaller of our two denominators.

This gives us , so .

Therefore, we can see that

.

The hyperbola pictured is centered at , meaning that the equation has a horizontal shift. The equation must have rather than just x. The hyperbola opens up and down, so the equation must be the y term minus the x term. The hyperbola is drawn according to the box going up/down 5 and left/right 2, so the y term must be or , and the x term must be or .

First, we must determine if the major axis is a vertical axis or a horizontal axis. We look at our denominators, and , and see that the larger one is under the -term. Therefore, we know that the greater axis will be a vertical one.

To find out how far the end point are from the center, we simply take . So we know the end points will be units above and below our center. To find the center, we must remember that for ,

the center will be .

So for our equation, the center will be . units above and below the center give us and .

In order to find the endpoints of the major and minor axes of our ellipse, we must first remember what each part of the equation in standard form means:

The point given by (h,k) is the center of our ellipse, so we know the center of the ellipse in the problem is (8,-2), and we know that the end points of our major and minor axes will line up with the center either in the x or y direction, depending on the axis. The parts of the equation that will tell us the distance from the center to the endpoints of each axis are and . If we take the square root of each, a will give us the distance from the center to the endpoints in the positive and negative x direction, and b will give us the distance from the center to the endpoints in the positive and negative y direction:

Now it is important to consider the definition of major and minor axes. The major axis of an ellipse is the longer one, will the minor axis is the shorter one. We can see that b=5, which means the axis is longer in the y direction, so this is the major axis. To find the endpoints of the major axis, we'll go 5 units from the center in the positive and negative y direction, respectively, giving us:

Similarly, to find the endpoints of the minor axis, we'll go 2 units from the center in the positive and negative x direction, respectively, giving us:

In order to find the endpoints of the major and minor axes of our ellipse, we must first remember what each part of the equation in standard form means:

The point given by (h,k) is the center of our ellipse, so we know the center of the ellipse in the problem is (8,-2), and we know that the end points of our major and minor axes will line up with the center either in the x or y direction, depending on the axis. The parts of the equation that will tell us the distance from the center to the endpoints of each axis are and . If we take the square root of each, a will give us the distance from the center to the endpoints in the positive and negative x direction, and b will give us the distance from the center to the endpoints in the positive and negative y direction:

Now it is important to consider the definition of major and minor axes. The major axis of an ellipse is the longer one, will the minor axis is the shorter one. We can see that b=5, which means the axis is longer in the y direction, so this is the major axis. To find the endpoints of the major axis, we'll go 5 units from the center in the positive and negative y direction, respectively, giving us:

Similarly, to find the endpoints of the minor axis, we'll go 2 units from the center in the positive and negative x direction, respectively, giving us:

First, we must determine if the major axis is a vertical axis or a horizontal axis. We look at our denominators, and , and see that the larger one is under the -term. Therefore, we know that the greater axis will be a vertical one.

To find out how far the end point are from the center, we simply take . So we know the end points will be units above and below our center. To find the center, we must remember that for ,

the center will be .

So for our equation, the center will be . units above and below the center give us and .

The hyperbola pictured is centered at , meaning that the equation has a horizontal shift. The equation must have rather than just x. The hyperbola opens up and down, so the equation must be the y term minus the x term. The hyperbola is drawn according to the box going up/down 5 and left/right 2, so the y term must be or , and the x term must be or .