Steply

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

Step by step, with the reason for every step · Conics from the general form · Precalculus

Step by step

The question

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

x2 + y2 + 2x − 4y − 4 = 0

1 Name the conic

Look at the squares.

x² and y² both have 1 in front.

So it is a circle.

only x² or only y² → parabola
x² and y², same number in front → circle ✓
same signs, different numbers → ellipse
different signs → hyperbola

2 Group x and y

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

(x2 + 2x) + (y2 − 4y) − 4 = 0

Move the plain number to the right: − 4 becomes + 4.

(x2 + 2x) + (y2 − 4y) = 4

3 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.

(x2 + 2x + 1) + (y2 − 4y) = 4 + 1
2 ÷ 2 = 1, 1 × 1 = 1

Careful!

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

Then the two sides stay equal.

Careful: add 1 on both sides

4 Complete the square for y

Half of −4 is −2.

(−2) times (−2) is 4.

We add 4 inside the bracket.

So we add 4 on the right too.

(x2 + 2x + 1) + (y2 − 4y + 4) = 4 + 1 + 4
−4 ÷ 2 = −2, (−2) × (−2) = 4

5 Write as squares

Each bracket is now a full square.

On the right, 4 + 1 + 4 is 9.

(x + 1)2 + (y − 2)2 = 9

It is the circle rule: center (−1, 2), radius 3.

circle: center (−1, 2), r = 3

6 Check

Let's check with a point on it: (2, 2), the right end of the circle.

Put it in the first equation.

2² + 2² + 2 × 2 − 4 × 2 − 4
= 4 + 4 + 4 − 8 − 4 = 0

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: complete the square for x.

Then: complete the square for y.

Then: write as squares.

Last: check.

Answer
(x + 1)2 + (y − 2)2 = 9

Your turn

Try this one yourself, then check:

Name the conic and write in standard form x2 + y2 + 6x − 6y + 9 = 0
Show the answer
(x + 3)2 + (y − 3)2 = 9

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