Recall that the standard formula of a hyperbola can come in two forms:

 and

, where the centers of both hyperbolas are .

When the term with  is first, that means the foci will lie on a horizontal transverse axis.

When the term with  is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at  and .

For a hyperbola with a vertical transverse access, the foci will be located at  and .

For the given hypebola in the question, the transverse axis is vertical and its center is located at .

Next, find .

The foci are then located at  and .

To find the foci of a hyperbola, first determine a and b, and then use the relationship

In this case, the major axis is horizontal since x comes first, so and .

Solve for c: add 9 to both sides 

take the square root

Since the center is and the major axis is the horizontal one, our foci are . The only choice that works is

.

To solve, simply use the follow equation where c is the length of the foci.

In this particular case,

Thus,

The standard form of the equation for a hyperbola is given by

The foci are located at (h+c, k) and (h-c, k), where c is found by using the formula

Since our equation is already in standard form, you can see that

, 

Plugging into the formula

So the foci are found at

 AND

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 graph should look like this:

Rotated 90 degrees, this graph would be opening up and down instead of left and right, so the equation will have the y term minus the x term.

The box that the hyperbola is drawn around will also rotate. It will now be up/down 2 and left/right 3.

This makes the correct equation

.

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 .

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

 .

In order to determine which direction the parabola opens, we must first put the equation in standard form, which can be expressed in one of the following two ways:

If the equation is for  as in the first above, the parabola opens up if  is positive and down if is negative. If the equation is for as in the second above, the parabola opens right if is positive and left if is negative. Rearranging our equation, we get:

We can see that our equation is for , which means the parabola will open either left or right. The sign of the first term is negative, so this parabola will open to the left.