What is the difference in these two equations?

[darshanpatel]

Homework Statement

What is the difference in these equations?

1. 4(x-1)^2)+(y+2)^2=0

2. (((x-1)^2)/4)+(((y+2)^2)/16)=1

3. ((-2(x-1)^2)/1)-(((y+2)^2)/2)=1

Homework Equations

-None-

The Attempt at a Solution

If you are wondering where i got these equations from where by changing the constant to +8, and +10 from -8 for this equation:

4x^2+y^2-8x+4y-8=0

I think the first one is a circle because it kind of looks like x^2+y^2=r^2, not sure..
The second one is just a basic ellipse because it follows its standard form.
The third one is a hyperbola because it is x^2-y^2

 

[Mark44]

Homework Statement

What is the difference in these equations?

1. 4(x-1)^2)+(y+2)^2=0

2. (((x-1)^2)/4)+(((y+2)^2)/16)=1

3. ((-2(x-1)^2)/1)-(((y+2)^2)/2)=1

Homework Equations

-None-

The Attempt at a Solution

If you are wondering where i got these equations from where by changing the constant to +8, and +10 from -8 for this equation:

4x^2+y^2-8x+4y-8=0

I think the first one is a circle because it kind of looks like x^2+y^2=r^2, not sure..

Nope, it's not a circle. If it were 4x² + y² = 1, what would it be?

How about if it were 4x² + y² = 0?

The second one is just a basic ellipse because it follows its standard form.

Yes.

The third one is a hyperbola because it is x^2-y^2

No it isn't. And you don't have x² - y² = 1, which is a hyperbola. Look closer at the signs.

 

[darshanpatel]

That means the first one is a 'v' graph? Do they have a certain name?

And the third one is a parabola.

 

[Mark44]

That means the first one is a 'v' graph?

No. Think about what the graph of 4(x-1)^2+(y+2)^2=0 looks like, and in particular, think about what happens to a number when you square it. The graph is very simple.

Do they have a certain name?

And the third one is a parabola.

No again. In the equation of a parabola there will be an x² term and no y² term, OR a y² term but no x² term. In this equation --
((-2(x-1)^2)/1)-(((y+2)^2)/2)=1

-- there are both an x² term and a y² term. Simplify the equation a bit and see if you can graph it.

 

[darshanpatel]

Didn't you simplify the first one already, because the x-1 and y+2 are just origin shifts so it would look like:

4x^2+y^2=0

I am thinking parabola...

The second one would be:

(-2x^2)-(y^2/2)=1

 

[Mark44]

Didn't you simplify the first one already, because the x-1 and y+2 are just origin shifts so it would look like:

4x^2+y^2=0

If you translate this graph 1 unit right and 2 units down, you get the graph of your first equation.

I am thinking parabola...

No, it can't be a parabola, because there is an x² term AND a y² term. Parabolas never have two second-degree terms.

If the equation were 4x² + y² = 1, what would it be?

What would the untranslated equation 4x² + y² = 0 be? What can you put in for x and y so that you get 0 on the right side?

The second one would be:

(-2x^2)-(y^2/2)=1

Well, all you did here was remove the translations. There's one more simplification that you can do.