Step by step
The question
Question: find the vertices, the asymptotes and the foci of (x − 3)² over 4 − y² over 4 = 1.
1 Find the center
The center comes from the brackets.
The sign turns: x − 3 gives 3, y alone gives 0.
So the center is (3, 0).
2 Find a and b
a² is the bottom under the plus part, the one with no minus.
It is under x.
So the curves open left and right.
a² = 4, so a = 2.
The other bottom is b² = 4, so b = 2.
3 Vertices
The vertices are where the curves start: a = 2 from the center, left and right.
Draw the grid.
Put the center on the grid.
And the two vertices.
4 Draw the box
A helper box: 2 each way across, and 2 each way up and down.
5 Asymptotes
The asymptotes go through the corners of the box.
The curves get closer and closer to them, but never touch.
Their slope is up over across of the box: 2/2, which is 1.
One goes up, one goes down.
So the asymptotes are y = (x − 3) and y = −(x − 3): the point-slope form, through the center.
Now the two curves: each starts at a vertex and bends out along the asymptotes.
6 Find the foci
The foci sit inside the curves, c from the center: c² = a² + b².
Careful!
For a hyperbola it is plus: the foci are further out than the vertices.
Minus is the ellipse rule.
Now put the numbers in: a² = 4 and b² = 4.
4 plus 4 is 8.
So c is the square root of 8.
But 8 has a square number in it: 4.
8 is 4 times 2.
A root of a times splits into two roots.
√4 is 2.
So c = 2√2.
That is the exact answer, made simple.
Go c from the center, the way the curves open: (3 − 2√2, 0) and (3 + 2√2, 0).
Put the foci on the grid, inside the curves.
7 Check
Let's check with a vertex, (5, 0).
Put it in the standard form.
We get 1, the right side.
So the vertex is on the hyperbola.
Right!
Let's look back.
First: find the center.
Then: find a and b.
Then: vertices.
Then: draw the box.
Then: asymptotes.
Then: find the foci.
Last: check.