Worked examples
The question
Question: find the vertices, the asymptotes and the foci of x² over 16 − y² over 9 = 1.
x216 − y29 = 1
1 Find the center
There are no brackets: just x² and y².
So the center is (0, 0).
center = (0, 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 is under the plus part: x → opens left and right
a² = 16, so a = 4.
a2 = 16 → a = 4
The other bottom is b² = 9, so b = 3.
b2 = 9 → b = 3
3 Vertices
The vertices are where the curves start: a = 4 from the center, left and right.
vertices = (−4, 0) and (4, 0)
Draw the grid.
Put the center on the grid.
And the two vertices.
4 Draw the box
A helper box: 4 each way across, and 3 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: 3/4.
One goes up, one goes down.
slope = 34 (up over across)
So the asymptotes are y = (3/4)x and y = −(3/4)x.
y = 34x and y = −34x
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.
c2 = a2 + b2
Careful!
For a hyperbola it is plus: the foci are further out than the vertices.
Minus is the ellipse rule.
Careful: not c² = 16 − 9 ✗
Now put the numbers in: a² = 16 and b² = 9.
c2 = 16 + 9
16 plus 9 is 25.
c2 = 25
5 × 5 = 25
So c = 5, because 5 times 5 is 25.
c = 5
Go c from the center, the way the curves open: (−5, 0) and (5, 0).
foci = (−5, 0) and (5, 0)
Put the foci on the grid, inside the curves.
7 Check
Let's check with a vertex, (4, 0).
Put it in the standard form.
(4)216 − (0)29 = 1616 − 09 = 1 − 0 = 1
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.
Answer vertices = (−4, 0) and (4, 0)
▶ Watch it solved
The question
Question: find the vertices, the asymptotes and the foci of y² over 9 − x² over 16 = 1.
y29 − x216 = 1
1 Find the center
There are no brackets: just x² and y².
So the center is (0, 0).
center = (0, 0)
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 is under the plus part: y → opens up and down
a² = 9, so a = 3.
It is not the bigger bottom here: for a hyperbola, the plus part decides.
a2 = 9 → a = 3
The other bottom is b² = 16, so b = 4.
b2 = 16 → b = 4
3 Vertices
The vertices are where the curves start: a = 3 from the center, down and up.
vertices = (0, −3) and (0, 3)
Draw the grid.
Put the center on the grid.
And the two vertices.
4 Draw the box
A helper box: 3 each way up and down, and 4 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: 3/4.
One goes up, one goes down.
slope = 34 (up over across)
So the asymptotes are y = (3/4)x and y = −(3/4)x.
y = 34x and y = −34x
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.
c2 = a2 + b2
Careful!
For a hyperbola it is plus: the foci are further out than the vertices.
Minus is the ellipse rule.
Careful: not c² = 16 − 9 ✗
Now put the numbers in: a² = 9 and b² = 16.
c2 = 9 + 16
9 plus 16 is 25.
c2 = 25
5 × 5 = 25
So c = 5, because 5 times 5 is 25.
c = 5
Go c from the center, the way the curves open: (0, −5) and (0, 5).
foci = (0, −5) and (0, 5)
Put the foci on the grid, inside the curves.
7 Check
Let's check with a vertex, (0, 3).
Put it in the standard form.
(3)29 − (0)216 = 99 − 016 = 1 − 0 = 1
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.
Answer vertices = (0, −3) and (0, 3)
▶ Watch it solved
The question
Question: find the vertices, the asymptotes and the foci of (x − 2)² over 9 − (y + 1)² over 16 = 1.
(x − 2)29 − (y + 1)216 = 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).
center = (2, −1)
x − 2 → h = 2
y + 1 → k = −1
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 is under the plus part: x → opens left and right
a² = 9, so a = 3.
It is not the bigger bottom here: for a hyperbola, the plus part decides.
a2 = 9 → a = 3
The other bottom is b² = 16, so b = 4.
b2 = 16 → b = 4
3 Vertices
The vertices are where the curves start: a = 3 from the center, left and right.
vertices = (−1, −1) and (5, −1)
Draw the grid.
Put the center on the grid.
And the two vertices.
4 Draw the box
A helper box: 3 each way across, and 4 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: 4/3.
One goes up, one goes down.
slope = 43 (up over across)
So the asymptotes are y + 1 = (4/3)(x − 2) and y + 1 = −(4/3)(x − 2): the point-slope form, through the center.
y + 1 = 43(x − 2) and y + 1 = −43(x − 2)
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.
c2 = a2 + b2
Careful!
For a hyperbola it is plus: the foci are further out than the vertices.
Minus is the ellipse rule.
Careful: not c² = 16 − 9 ✗
Now put the numbers in: a² = 9 and b² = 16.
c2 = 9 + 16
9 plus 16 is 25.
c2 = 25
5 × 5 = 25
So c = 5, because 5 times 5 is 25.
c = 5
Go c from the center, the way the curves open: (−3, −1) and (7, −1).
foci = (−3, −1) and (7, −1)
Put the foci on the grid, inside the curves.
7 Check
Let's check with a vertex, (5, −1).
Put it in the standard form.
(5 − 2)29 − (−1 + 1)216 = 99 − 016 = 1 − 0 = 1
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.
Answer vertices = (−1, −1) and (5, −1)
▶ Watch it solved
