Step by step
The question
Question: find the vertices, the asymptotes and the foci of 4x² − y² = 4.
1 Make the right side 1
The standard form has 1 on the right.
So we divide every part by 4.
We do it on both sides, so they stay equal.
4/4 is 1/1.
So x² − y² over 4 = 1.
2 Find the center
There are no brackets: just x² and y².
So the center is (0, 0).
3 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 = 1, so a = 1.
It is not the bigger bottom here: for a hyperbola, the plus part decides.
The other bottom is b² = 4, so b = 2.
4 Vertices
The vertices are where the curves start: a = 1 from the center, left and right.
Draw the grid.
Put the center on the grid.
And the two vertices.
5 Draw the box
A helper box: 1 each way across, and 2 each way up and down.
6 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: 2/1.
One goes up, one goes down.
So the asymptotes are y = 2x and y = −2x.
Now the two curves: each starts at a vertex and bends out along the asymptotes.
7 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 = 1 and b² = 4.
1 plus 4 is 5.
So c is the square root of 5.
5 is not a square number.
So we keep √5.
That is the exact answer.
It is about 2.2.
Go c from the center, the way the curves open: (−√5, 0) and (√5, 0).
Put the foci on the grid, inside the curves.
8 Check
Let's check with a vertex, (1, 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: make the right side 1.
Then: find the center.
Then: find a and b.
Then: vertices.
Then: draw the box.
Then: asymptotes.
Then: find the foci.
Last: check.