Graph 4y^2 - x^2 = 16 Step by Step

Step by step

The question

Question: find the vertices, the asymptotes and the foci of 4y² − x² = 16.

4y² − x² = 16

1 Make the right side 1

The standard form has 1 on the right.

So we divide every part by 16.

We do it on both sides, so they stay equal.

4y²16 − x²16 = 1616

4/16 is 1/4.

16 ÷ 4 = 4, 16 ÷ 1 = 16

So y² over 4 − x² over 16 = 1.

y²4 − x²16 = 1

2 Find the center

There are no brackets: just x² and y².

So the center is (0, 0).

center = (0, 0)

3 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 = 4, so a = 2.

It is not the bigger bottom here: for a hyperbola, the plus part decides.

a² = 4 → a = 2

The other bottom is b² = 16, so b = 4.

b² = 16 → b = 4

4 Vertices

The vertices are where the curves start: a = 2 from the center, down and up.

vertices = (0, −2) and (0, 2)

Draw the grid.

Put the center on the grid.

And the two vertices.

5 Draw the box

A helper box: 2 each way up and down, and 4 each way across.

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: 2/4, which is 1/2.

One goes up, one goes down.

slope = 24 (up over across)

So the asymptotes are y = (1/2)x and y = −(1/2)x.

y = 12x and y = −12x

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.

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 − 4 ✗

Now put the numbers in: a² = 4 and b² = 16.

c² = 4 + 16

4 plus 16 is 20.

c² = 20

So c is the square root of 20.

c = √20

But 20 has a square number in it: 4.

20 is 4 times 5.

c = √4 × 5

A root of a times splits into two roots.

c = √4 × √5

√4 is 2.

2 × 2 = 4

So c = 2√5.

That is the exact answer, made simple.

c = 2√5

Go c from the center, the way the curves open: (0, −2√5) and (0, 2√5).

foci = (0, −2√5) and (0, 2√5)

Put the foci on the grid, inside the curves.

8 Check

Let's check with a vertex, (0, 2).

Put it in the standard form.

(2)²4 − (0)²16 = 44 − 016 = 1 − 0 = 1

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.

Answer