Is there a way of finding the equation of a hyperbola without having to draw a table?

Question

Sorry. I'm new to hyperbola's. What I'm trying to say is this. I've been given several hyperbolas and have to match them up with their equation. Is their a method of finding the equation for that hyperbola straight out. For example, if you had a parabola you could work out the equation by finding the x and y intercepts, vertex and axis of symmetry.

I'm serious. I don't want to cheat by going on Yahoo Answers. I'd like to know how to do it and if it's possible. Thanks!

What I mean is that you've been given a hyperbola like the one in the link below and have been asked to work out the equation of the hyperbola.

http://hotmath.com/images/gt/lessons/genericalg1/hyperbola.gif

Answer 1

You don't say whether the hyperbolas all have an axis parallel to the y or x axis. If they don't, then in general, finding the equation is a lot more difficult.

A hyperbola centred at the origin has the equation:
x^2 / a^2 - y^2 / b^2 = 1 if it is EW-opening, or
y^2 / a^2 - x^2 / b^2 = 1 if it is NS-opening.
The negative sign determines which way the hyperbola opens.
If the asymptotes are perpendicular to each other, then a = b in the formulae above.

If an EW-opening hyperbola is centered at (h,k) instead of at (0,0), its equation becomes:
(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1.
Its centre is (h, k).
The foci are (h + c, k) and (h - c, k)
where c = sqrt(a^2 + b^2).
Its asymptotes are:
(x - h)^2 / a^2 = (y - k)^2 / b^2
(x - h) / a = +/- (y - k) / b.

At the vertices, y = k.
Therefore:
(x - h)^2 / a^2 = 1
x = h +/- a
The vertices are (h - a, k) and (h + a, k).

Depending on what information you are given, you have to use formulae similar to these to work back to the equation.

In the diagram you added later, the hyperbola is a rectangular one centered at the origin. The axes have been rotated through 45 degrees to become the asymptotes. For a hyperbola in this position, the equation is:
xy = c^2.

You can see from your diagram, that y = 1 when x = 1, and that means c^2 = 1, making the equation:
xy = 1 or
y = 1/x.

If the hyperbola were centred at (h,k) instead of (0,0), its equation would be:
(x - h)(y - k) = c^2.

Answer 2

The answers above are good for hyperbolas that align themselves with the x and y axes (i.e. that open vertically or horizontally). However the example at the link you gave was of a pair of hyperbolas at 45° angles to the x and y axes. When the angle is 45° the equation has a relatively simple form. For such a hyperbola with center (h, k) we have:

y - k = a/(x - h)

In this particular case the hyperbola is centered on the origin. So the equation becomes

y = a/x

Now you need to solve for a.

Plug in a point on the hyperbola. Let's choose (1, 1).

y = a/x
1 = a/1
a = 1

So the equation is:

y = 1/x

Answer 3

A hyperbola is a conic section and it has a standard form. I don't understand the question.

The standard cartesian form of a hyperbola is

(x-a)^2/m^2 - (y-b)^2/n^2 = 1

Okay. If the equations you're given are in the form I mentioned, then the centroid of the hyperbola will be at the point (a,b) and the assymptotes will have slope +/-m/n if the coefficient of the y^2 term is negative and +/-n/m if the coefficient of the x^2 term is negative, in other words, depending upon whether the hyperbola opens out to the left and right or up and down. It's a bit hard to explain without a picture.

Answer 4

A hyperbola with center at (h,k) has the equation:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1 [transverse axis || x-axis], and,
(y-k)^2/a^2 - (x-h)^2/b^2 =1 [transverse axis || y-axis].

2a is the length of the transverse axis which is the distance between the vertices of the the hyperbola.

The foci are located on the transverse axis at a distance c from the vertex. The distance between the two foci = 2c.

b^2 = c^2-a^2

The eccentricity = c/a and is always > 1

the asymptotes are given by y=+/- bx/a

Answer 5

You can READ graphic information directly from the equation.

(x-h)² . (y-k)²
------- - ------ = 1
.. a² ..... b²

the center of the hyperbola is (h,k)
the minus in front of the y term tells us the hyperbola opens left and right. The vertices are at (h±a, k). Slope of the asymptotes is ±b/a.

If the minus had been in front of the x term, the hyperbola would have opened up and down and the vertices would have been (h, k±b). There's more, but that's why they write chapters in math books.

Answer 6

What do you mean by- "you are given some hyperbolas"?

If they gave you equation, what more do you want?

If they gave you figures, choose some six points,
substitute and solve a simultaneous equation.

You can test if it is a conic by doing pascal's test ;) on
vertices.

Answer 7

you need atleast two zeros (solutions) in order to figure out the equation of a parabola.

Answer 8

Yes, go to Yahoo! Answers and get someone to do it for you.