unsure? + hyperbola!

**frogsrcool** · November 22nd 2006, 03:25 PM

okay how would I go about and do this question?
determien the standard form of the equation of a hyperbola with vertices(+2,0) and passing through (4,3)

would I use the formula:
(y-k)^2/a^2 -(x-h)^2/ b^2

but what would a and b stand for??

**galactus** · November 22nd 2006, 03:52 PM

An equation of the hyperbola has the form:

$\frac{x^{2}}{2^{2}}-\frac{y^{2}}{b^{2}}=1$

Since P(4,3) passes through, the x and y coordinates satisfy the equation:

$\frac{4^{2}}{2^{2}}-\frac{3^{2}}{b^{2}}=1$

Solving, we find $b^{2}=3$

So, the equation is:

$\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$

**TriKri** · November 22nd 2006, 03:55 PM

Take a look at this site for better understanding.

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2} = 1$ will give you an north-south opening hyperbola. $(h, k)$ is the center of the hyperbola. a is the closest the hyperbola ever gets to the center. And b/a gives the absolute value of the ratio x/y for points infinetly far away from the center. Was that helpful?

Now I don't know what a vertice is, but you have four unknown, a, b, k and h. You got to have at least four equations to be able to get the values of those unknown.

You have for example one pair of coordinates. You can always insert them to the equation. So, since you got
$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2} = 1$
and you know that (x, y) = (4, 3) is one point in the hyperbola, you can as a beginning set up the equation
$\frac{(3-k)^2}{a^2}-\frac{(4-h)^2}{b^2} = 1$

Good luck!

**topsquark** · November 22nd 2006, 05:29 PM

Originally Posted by frogsrcool

okay how would I go about and do this question?
determien the standard form of the equation of a hyperbola with vertices(+2,0) and passing through (4,3)

would I use the formula:
(y-k)^2/a^2 -(x-h)^2/ b^2

but what would a and b stand for??

Originally Posted by TriKri

Now I don't know what a vertice is

From reading the question I would assume that frogsrcool used the word "vertex" instead of the more standard term "focus."

-Dan

**galactus** · November 23rd 2006, 01:59 AM

Yes, there are vertices on a hyperbola. The vertices are the x-intercepts.

Thay would have coordinates V(a,0) and V(-a,0). This one has vertices
V(2,0) and V'(-2,0).

Line V'V is called the transverse axis. The end points of this line segment are the vertices.

The foci of this particular hyperbola would be:

$c^{2}=a^{2}+b^{2}=4+3=7$

The foci are at $(\pm\sqrt{7},0)$

**topsquark** · November 23rd 2006, 02:11 AM

Originally Posted by galactus

Yes, there are vertices on a hyperbola. The vertices are the x-intercepts.

Thay would have coordinates V(a,0) and V(-a,0). This one has vertices
V(2,0) and V'(-2,0).

Line V'V is called the transverse axis. The end points of this line segment are the vertices.

The foci of this particular hyperbola would be:

$c^{2}=a^{2}+b^{2}=4+3=7$

The foci are at $(\pm\sqrt{7},0)$

My apologies. I have never heard of the term being applied to an hyperbola before.

-Dan

**galactus** · November 23rd 2006, 03:19 AM

Certainly no need to apologize. I feel honored that I showed you something. I went ahead and included a graph of the said hyperbola.

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