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:

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:

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:

Since the x-coordinates of each foci are the same, this hyperbola is transverse axis vertical. The center is the midpoint of the foci, which is  in this case. The a and c values are found by calculating the distance between the center and vertices and foci, respectively.

In this problem,  and . The b value is found through the Pythagoreom Theorem, .

Plugging the values for the center  and a, b, and c, the following equation can be used to find the standard form of the parabola:

To put this into standard form, we will have to complete the square for both x and y. To do this, it will be really helpful to re-group our terms:

Factor the -2 out of the y terms:

Adding 4 completes the square for x, since

Adding 9 completes the square for y, since

We're adding this inside the parentheses, so really we are adding to both sides

re-write the left side with squared binomials, and add the numbers on the right side

divide both sides by 17

To write this in standard form, first we will have to complete the square for both x and y. It will be helpful to re-group the terms:

 Factor out the coefficients for both x squared and y squared

Adding 1 completes the square for x since

Since we're adding it inside those parentheses, we're really adding

Adding 16 completes the square for y since

Since we're adding it inside those parentheses, we're really adding

divide both sides by 6

To write this in standard form, first complete the square for x and y. It will be helpful to re-group our terms:

factor out the coefficients on the squared terms

Adding 9 will complete the square for x since

We're adding it in the parentheses, so really we're adding to both sides

Adding 1 will complete the square for y since

We're really adding to both sides

divide both sides by 36

The given clue is that the hyperbola will open either up or down.  This indicates that the regular  and  variables are swapped.  

If the hyperbola opens left or right, the equation of the hyperbola is:

Write the form of a hyperbola in the second form.

Substitute the center .

The hyperbola must minimally be in this form.  

The correct answer is:  

Re-write the equation so that the x terms are next to each other:

factor our 4 from the x terms, and add 20 to both sides

complete the square for x by adding 1 inside the parentheses. To keep both sides equal, add 4 to the right side as well, since the 4 is really being distributed to the 1 we added on the left.

re-write the x terms as a binomial squared

divide both sides by 24

To be written in standard form, the equation of a hyperbola must look like one of the following:

     OR     

Our equation looks most like the one on the left, since the x term is first.  To obtain standard form, we must have 1 on the right side of the equation.  To do this, we divide both sides of the equation by 128. 

Now, we can simplify the fractions to obtain standard form