To be fully generalized, an hyperbola's equation is as follows:

A*x*^{2} + B*xy* + C*y*^{2} + D*x* + E*y* + F = 0

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You probably noticed that B*xy* term, which has not been present in any of the hyperbola equations you've seen thus far. This is because the B*xy* term is present in the vast majority of hyperbolas that do *not* parallel either of the coordinate axes. In other words, you are, in your pre-calculus context, dealing only with a specific subset of all hyperbolas, being the ones with side-by-side or top-and-bottom branches.

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Which is good, because all those other "angled" hyperbolas will require (in calculus) that you do what's called "rotation of axes" to work with them, and that process is a royal pain. Plus it requires trigonometry, which this lesson assumes you haven't yet taken. Here, you'll just be working with the nice hyperbolas.

(Yes, I still remember having to do this thirty years ago in calculus. No, I still haven't gotten over it.)

There seems to be no consensus on what the "general" or "standard" form of an hyperbola's equation is; I've seen the conics form referred to as the standard form. So I'll try to be clear: conics, vertex, or center-vertex form is the squared-variable, equals-one form; the other form is the multiplied-out form. You'll need to pay close attention in your class to the terms used by your textbook and instructor, so you can be sure of the meaning for the homework and the test.

- Find the center-vertex equation for the hyperbola with center (2, 3), vertex (0, 3), and focus (5, 3).

The center, focus, and vertex all lie on the horizontal line *y* = 3 (that is, they're side by side on a line paralleling the *x*-axis), so the branches must be side by side, and the *x* part of the equation must be what's added. The *a*^{2} will go under the *x*^{2}.

The *y* part then will be subtracted, and the *b*^{2} will go under the *y*^{2}.

Looking at the coordinates of the center point, I can see that the squared-variable parts of my equation will be (*x* − 2)^{2} and (*y* − 3)^{2}.

Comparing the center point and the vertex point, I see that the given vertex is 2 units from the center, so *a* = 2; the given focus is 3 units from the center, so *c* = 3. Then:

*b*^{2} = (3)^{2} − (2)^{2} = 5

(I don't need to bother with the value of *b* itself, since they only asked me for the equation, in which I'll be using the square of *b*.)

Putting this all together, my equation is:

- Find the multiplied-out equation for the hyperbola with center (0, 0), vertex (0, 5), and asymptotes

The vertex and the center are both on the vertical line *x* = 0 (that is, on the *y*-axis), so the hyperbola's branches are above and below each other, not side by side.

Then the *y* part of the equation will be added, and will get the *a*^{2} as its denominator.

From the coordinates of the center (being the origin), I can see that the squared variable parts will be simply *x*^{2} and *y*^{2}.

(The slopes of the two asymptotes are of the form . It looks like *a* will be 5 and *b* will be 3, but I'll derive the values directly, in case the slope value represents a fraction that was reduced.)

The vertex they gave me is 5 units above the center, so *a* = 5 and *a*^{2} = 25. Then slope of the asymptotes (ignoring the "plus-minus" part) is:

*b* = 3

*b*^{2} = 9

And this is all I need in order to find my equation:

Checking the instructions, I see that they want the multiplied-out form of the equation, so I'll multiply through by 25×9 to get my answer:

9*y*^{2} − 25*x*^{2} = 225

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- Find the center-vertex equation of the hyperbola with
*x*-intercepts at*x*= −5 and*x*= 3, and foci at (−6, 0) and (4, 0).

The foci are side by side, so this hyperbola's branches are side by side, and the center, foci, and vertices lie on a line paralleling the *x*-axis. So the *y* part of the equation will be subtracted and the *a*^{2} will go with the *x* part of the equation.

The center is midway between the two foci, so the center must be at (*h*, *k*) = (−1, 0). This tells me that the squared-variable parts will be (*x* −(−1))^{2} = (*x* + 1)^{2} and (*y* − 0)^{2} = *y*^{2}.

The foci are 5 units to either side of the center, so *c* = 5 and *c*^{2} = 25.

The center lies on the *x*-axis, so the two *x*-intercepts must then also be the hyperbola's vertices. Since the intercepts are 4 units to either side of the center, then *a* = 4 and *a*^{2} = 16. Then:

*a*^{2} + *b*^{2} = *c*^{2}

*b*^{2} = 25 − 16 = 9

Then my equation is:

- Find the vertex equation for the hyperbola with vertices at (−2, 15) and (−2, −1), and having eccentricity

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The vertices are above and below each other, so the center, foci, and vertices lie on a vertical line paralleling the *y*-axis. Then the *a*^{2} will go with the *y* part of the hyperbola equation, and the *x* part will be subtracted.

The center is midway between the two vertices, so (*h*, *k*) = (−2, 7). This tells me that the squared-variable parts of the equation will be (*x* − (−2))^{2} = (x + 2)^{2} and (*y* − 7)^{2}.

The vertices are 8 units above and below the center, so *a* = 8 and *a*^{2} = 64.

The eccentricity is , so *c* = 17 and *c*^{2} = 289. Then:

(8)^{2} + *b*^{2} = 289

*b*^{2} = 289 − 64 = 225

This is all I need in order to create my equation, which is:

- A cooling tower's cross-section is modelled by an hyperbola. The base of the tower is 520 feet (ft) wide. The center of the modelling curve is 360 ft above the base. The tower, at its narrowest point, is 200 ft wide; the top of the tower is 200 ft above this narrowest point. Find an equation of the hyperbola. From this equation, find the width of the opening at the tower's top.

A cooling tower is vertical, so its sides (being the vertical cross-section) can be portions of the two branches of a side-by-side hyperbola.

For this modelling equation, the *y* part will be subtracted, and the *x* part will get the *a*^{2}.

I will center the model around the *y*-axis, with the midpoint of the base being at the origin. Then, because the center of the model is 360 ft above the base, the center of my equation will be at (*h*, *k*) = (0, 360).

The narrowest point of any hyperbola will be the distance between the vertices. The vertices in this case will be at the height of the center point, and will be 200 ÷ 2 = 100 ft to either side. This means that *a* = 100.

So far, I've got this much of my equation:

I still need to find the value of *b*. To do this, I'll plug in some other information from the exercise.

They tell me that the base is 520 ft wide. Then, from the center line (being the axis of symmetry; in this case, being the *y*-axis), the base crosses the hyperbola at a distance of 520 ÷ 2 = 260. So the line of the base intersects the hyperbola at the points (±260, 0). I'll plug this in — and the "plus-minus" won't matter, because I'll be squaring the *x*-value anyway — and solve for the value of *b*^{2}.

This then gives me my modelling hyperbola:

To find the width of the opening at the top of the tower, I'll need to plug in a point where the horizontal line for the top of the tower intersects the hyperbola's branches. Since the top of the tower is 200 ft above the narrowest point (and the narrowest point is at the height of the center point), then I get the intersection points being at height 360 + 200 = 560 ft above the base on the *x*-axis.

So I know that I have the point (*x*, 560) as the intersection point; more usefully, I know that this point is on the hyperbola, so I can plug it into my equation.

Since I'm looking for a distance, I'll use the "plus" part of the answer. This answer is the distance from the center-line of the hyperbola to the branches on either side; in this context, this answer is the radius of the circle formed by the top opening of the tower. They asked me for the diameter, which is twice the radius, so I multiply by 2 and get ft.

Converting this fraction to mixed-number form, I get a diameter of ft. The one-third of a foot is inches.

So my answer is:

top's diameter: 333 ft, 4 inches

Note that the instructions for this last exercise had said to find "an" equation for the hyperbola.

One exercise can have more than one hyperbola equation as its answer. Besides the options for format of the equation (namely, conics form or else multiplied-out form), there will be more than one possible equation for this hyperbola because of the options for how I choose to place my model in the plane. Will I have my hyperbola straddle the *y*-axis, or will I slide it to one side or the other, maybe putting the left-hand end of the base at the origin? Will I put the center of my equation at the origin, or put it somewhere above the *x*-axis? Either of these choices could have been made in a way other than I did above, and either would have generated just as valid an equation.

Precisely because there will be different answer options, make sure that you show your work and reasoning clearly, so that your grader can see where you were headed.

URL: https://www.purplemath.com/modules/hyperbola3.htm

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