Which conic is x^2 + 4y^2 - 2x + 8y + 1 = 0

Step by step

The question

Question: name the conic and write in standard form x² + 4y² − 2x + 8y + 1 = 0.

1 Name the conic

Look at the squares.

x² and y² have the same sign, but 1 and 4 are different numbers.

So it is an ellipse.

2 Group x and y

Put the x terms together in a bracket, and the y terms together.

Move the plain number to the right: + 1 becomes − 1.

3 Take out the number in front

Inside each bracket, x² and y² must have 1 in front.

So we take out 4.

4 Complete the square for x

Half of −2 is −1.

(−1) times (−1) is 1.

We add 1 inside the bracket.

So we add 1 on the right too.

Careful!

What we add on the left, we add on the right too.

Then the two sides stay equal.

5 Complete the square for y

Half of 2 is 1.

1 times 1 is 1.

We add 1 inside the bracket.

But the bracket is multiplied by 4.

So on the left we really added 4 times 1: 4.

We add 4 on the right too.

6 Write as squares

Each bracket is now a full square.

On the right, −1 + 1 + 4 is 4.

7 Make the right side 1

Divide every part by 4, so the right side is 1.

4/4 is 1/1.

That is the standard form of an ellipse with center (1, −1).

8 Check

Let's check with a point on it: (3, −1), a vertex.

Put it in the first equation.

We get 0, like the first equation says.

So the standard form is right!

Let's look back.

First: name the conic.

Then: group x and y.

Then: take out the number in front.

Then: complete the square for x.

Then: complete the square for y.

Then: write as squares.

Then: make the right side 1.

Last: check.