Hyperbola

That's a p.

Hyperbola

But a hyperbola would look something like this. They get to be right here and they get really close to the asymptote.

Asymptotes of a Hyperbola

They're the circle, the ellipse, the parabola, and the hyperbola.

Let's see if we can learn a thing or two about the hyperbola.

That an ellipse is just a squished circle.

And you'll get no solution.

We have it in standard form and, yes, indeed, we do have a hyperbola.

So the first thing we know is where the center of this hyperbola is.

And up here.

So these are both hyperbolas.

The two asymptotes of a hyperbola.

But I just wanted to make sure you realize that.

This cancels with this.

So if we go with this one, let's just solve for y.

Hyperbola by Asymptotes Point

So I'll do that right now.

And we've drawn our asymptotes.

Hyperbola by Focuses Point

Equilateral Hyperbola by Four Points

This is the asymptote.

Returns the inverse hyperbolic cosine of arg, i.e. the value whose hyperbolic cosine is arg.

Returns the inverse hyperbolic tangent of arg, i.e. the value whose hyperbolic tangent is arg.

And actually, just so you know, this equation right here, this is a particular hyperbola that opens to the left and the right.

Construct a hyperbola with this focus

left and right points on a hyperbola.

It actually doesn't take too long.

This looks like a really complicated thing.

Now this looks a lot like our original equation of the hyperbola, which was x squared over a squared minus y squared over b squared is equal to 1.

Construct a hyperbola through this point

It's above the asymptote there.

I have a feeling I might be running out of time.

But now this tells us the slope of the two asymptotes.

Returns the inverse hyperbolic sine of arg, i.e. the value whose hyperbolic sine is arg.

Select a point for the new hyperbola to go through...

Let me do it here actually, I want to do that other hyperbola.

So it's an x is equal to minus 1. y is equal to 2.

You go 1 2 3 minus 3, minus 1.

A hyperbola with given asymptotes through a point

Especially if it has the same asymptotes just shifted, but centered at 0 it would look like this x squared over 16 minus y squared over 4 is equal to 1.

So your distance between the two left and right points in a horizontal ellipse is the same as the distance between the two

An equilateral hyperbola constructed through four points

Let's add the hyperbola.