How to calculate the equation of a Hyperbola

  • I
  • Thread starter Dinoduck94
  • Start date
  • Tags
    Hyperbola
In summary, @.Scott found that there is no unique solution to the equation of a hyperbola, as it can be solved for in a variety of ways depending on the assumptions made.
  • #1
Dinoduck94
30
4
TL;DR Summary
How do you calculate the equation of a hyperbola, knowing only the following:

The y intercept: (0, y)
The area bounded by x=0: A
How do you calculate the equation of a hyperbola, knowing only that the y intercept is (0,y) and the area bounded at x=0 is 'A'.
It stands to reason that this can be calculated, but I can't find a tutorial or something similar online to help me answer this.

This isn't homework, I'm just trying to further my understanding of maths particularly in regards to cosmology; but I would appreciate it if it was treated as such. I want to understand the process of how the answer is reached.

If it helps for ease of explaining we can treat the y intercept as (0,-10) and the Area as 100.
 
Physics news on Phys.org
  • #2
Are you also assuming that the center of the hyperbola falls along the y axis? Or that the hyperbola is oriented so the lobes are pointed straight up and down? Or that the hyperbola has specific x/y proportions?

If not, there is no single solution.

Assuming only that it is vertically aligned:
## ((x−O_x)/S_x)^2−((y−O_y)/S_y)^2=1 ##
Where:
##S_x## and ##S_y## are the scale in the X and Y directions.
## (Ox,Oy) ## is the center.
 
Last edited:
  • #3
.Scott said:
Are you also assuming that the center of the hyperbola falls along the y axis? Or that the hyperbola is oriented so the lobes are pointed straight up and down? Or that the hyperbola has specific x/y proportions?

If not, there is no single solution.

Hi Scott,

The assumptions would be that the center of the hyperbola falls along the y axis.

Thank you
 
  • #4
And you are assuming that the hyperbola is vertically aligned.
So the equation are:
Assuming only that it is vertically aligned:
## ((y-O_y)/S_y)^2-(x/S_x)^2=1 ##
Where:
##S_x## and ##S_y## are the scale in the X and Y directions.
## (0,O_y) ## is the center.
So you have three unknowns (## O_y, S_x, S_y ##) and two inputs.
You still have one degree of freedom to account for.
 
Last edited:
  • #5
.Scott said:
And you are assuming that the hyperbola is vertically aligned.
So the equation are:
Assuming only that it is vertically aligned:
## (x/S_x)−((y-O_y)/S_y)=1 ##
Where:
##S_x## and ##S_y## are the scale in the X and Y directions.
## (0,O_y) ## is the center.
So you have three unknowns (## O_y, S_x, S_y ##) and two inputs.
You still have one degree of freedom to account for.

Sorry, I don't understand how the known area comes into all this. I appreciate that this is the standard equation for a vertical transverse axis - but how can this help with the information that is known?
 
  • #6
Actually, it [might be] solvable. (but it turns out there is no unique solution - see below)
Give me a moment...
 
Last edited:
  • Like
Likes Dinoduck94
  • #7
## y_i ##: The y-intercept - ##(0,y_i)## is upper point of lower hyperbolic lobe.
So the scale and the y-offset will be related this way:
## ((y_i-O_y)/S_y)^2 - (x/S_x)^2 = 1 ##
## ((y_i-O_y)/S_y)^2 - 0 = 1 ##
## y_i-O_y = +/- S_y ##
Scale is taken as positive...
## y_i = O_y - S_y ##

If we start with ##S_x = 1##, we get this:
## ((y-O_y)/S_y)^2 - (x/S_x)^2 = 1 ##
## (y-O_y)/S_y = (x^2 + 1)^{1/2} ##
## y = (S_y(x^2 + 1)^{1/2}) + O_y ##

At this point, I could calculate the +/- x bounds and then integrate to get the area. Then I could adjust the x scale to make it A, the target area.

But I am still able to select from any ##O_y, S_y## combination that satisfies ## y_i = O_y - S_y ##.

As I change the scale in the Y, I will get a different shape curve - but I will still be able to compensate with the X Scale to get area A.

There is no unique solution.
 
  • #8
Dinoduck94 said:
How do you calculate the equation of a hyperbola, knowing only that the y intercept is (0,y) and the area bounded at x=0 is 'A'.
The conclusion that @.Scott reached was that there is not a unique solution. From what I can see, the problem as stated is highly ambiguous.
What sort of hyperbola is it? Does the hyperbola open left and right or does it open up and down? In the first case, the formula of such a hyperbola centered at (0, 0) would be ##\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1##. In the second case, the corresponding formula would be ##\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1##.

What does "area bounded at x = 0" mean? The arms of a hyperbola extend out infinitely far, so if you want to describe an area, there need to be more lines that surround some area inside the arms of the hyperbola or under it and above/below some axis. As far as I can tell, we really don't have enough information to go on here.

For more information, see https://en.wikipedia.org/wiki/Hyperbola
 

1. How do I determine the center of a hyperbola?

The center of a hyperbola can be found by identifying the coordinates of the two foci and calculating the midpoint between them. This point will be the center of the hyperbola.

2. What is the equation for a horizontal hyperbola?

The equation for a horizontal hyperbola is (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where (h,k) is the center of the hyperbola and a and b are the distances from the center to the vertices along the x- and y-axes, respectively.

3. How do I find the foci of a hyperbola?

The foci of a hyperbola can be found by using the formula c^2 = a^2 + b^2, where c is the distance from the center to each focus and a and b are the distances from the center to the vertices along the x- and y-axes, respectively.

4. Can a hyperbola have a negative center?

Yes, a hyperbola can have a negative center. The coordinates of the center will depend on the specific equation of the hyperbola.

5. How many asymptotes does a hyperbola have?

A hyperbola has two asymptotes, which are lines that the hyperbola approaches but never touches. These asymptotes are found by using the formula y = ±(b/a)x, where a and b are the distances from the center to the vertices along the x- and y-axes, respectively.

Similar threads

Replies
3
Views
219
  • Calculus and Beyond Homework Help
Replies
8
Views
540
Replies
2
Views
1K
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • General Math
Replies
2
Views
770
Replies
3
Views
1K
Replies
7
Views
2K
  • Calculus
Replies
3
Views
2K
Back
Top