Step by step
The question
Question: find the vertices, the asymptotes and the foci of (y − 1)² over 4 − (x + 2)² over 9 = 1.
1 Find the center
The center comes from the brackets.
The sign turns: x + 2 gives −2, y − 1 gives 1.
So the center is (−2, 1).
2 Find a and b
a² is the bottom under the plus part, the one with no minus.
It is under y.
So the curves open up and down.
a² = 4, so a = 2.
It is not the bigger bottom here: for a hyperbola, the plus part decides.
The other bottom is b² = 9, so b = 3.
3 Vertices
The vertices are where the curves start: a = 2 from the center, down and up.
Draw the grid.
Put the center on the grid.
And the two vertices.
4 Draw the box
A helper box: 2 each way up and down, and 3 each way across.
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: 2/3.
One goes up, one goes down.
So the asymptotes are y − 1 = (2/3)(x + 2) and y − 1 = −(2/3)(x + 2): 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 = 9.
4 plus 9 is 13.
So c is the square root of 13.
13 is not a square number.
So we keep √13.
That is the exact answer.
It is about 3.6.
Go c from the center, the way the curves open: (−2, 1 − √13) and (−2, 1 + √13).
Put the foci on the grid, inside the curves.
7 Check
Let's check with a vertex, (−2, 3).
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.