Conics - How to identify figures in the plane?

[José Ricardo]

Homework Statement

Identify the following figures in the plane, saying if they are circles, ellipses, hyperboles or parabolas and giving their centers, rays, foci, and guidelines as the case may be:

Homework Equations

x² + y² - 3x = 0

The Attempt at a Solution

x² + y² - 3x = 0
x(x - 3) + y² = 0
c = (3,0)
(x-3)² + y² = 0
(3-3)² + y² -3 = 0
y = 0
x² + 0² - 3x = 0
x(x-3) = 0
x = 3
y = 0
It's a circle, the two square letters are round perfectly.

I would like you to know my errors.

 

[tnich]

Homework Statement

Identify the following figures in the plane, saying if they are circles, ellipses, hyperboles or parabolas and giving their centers, rays, foci, and guidelines as the case may be:

Homework Equations

x² + y² - 3x = 0

The Attempt at a Solution

x² + y² - 3x = 0
x(x - 3) + y² = 0
c = (3,0)
(x-3)² + y² = 0
(3-3)² + y² -3 = 0
y = 0
x² + 0² - 3x = 0
x(x-3) = 0
x = 3
y = 0
It's a circle, the two square letters are round perfectly.

I would like you to know my errors.

You are correct that it is a circle. You have not correctly found the center. For a circle, you should also find the radius. To do this, put the equation of the circle in this form:
$(x-a)^2 +(y-b)^2 = r^2$
This expresses the relationship that each point $(x,y)$ on the circle is a distance $r$ from the center $(a,b)$.

 

[member 587159]

From what I understand, you calculate points where the curve intersects with the x-axis.

This is of course not sufficient. If you want to show it is a circle, then show you can write it in the form

$(x-x_0)^2 + (y-y_0)^2 = r^2$

Hint: complete the square with respect to $y$

 

[vela]

x(x - 3) + y² = 0
(x-3)² + y² = 0

Your first mistake is this step: $(x-3)^2 \ne x(x-3)$.

 

[José Ricardo]

(x - 3)² + (y - 0)² = 0

 

[José Ricardo]

Your first mistake is this step: $(x-3)^2 \ne x(x-3)$.

Thanks a lot, Vela!

 

[member 587159]

(x - 3)² + (y - 0)² = 0

That can't be correct. A circle has a strictly positive radius.

Moreover, $(x-3)^2 + (y-0)^2 = x^2 + 9 - 6x + y^2$ is not what you were given.

 

[tnich]

(x - 3)² + (y - 0)² = 0

If you multiply out $(x-3)(x-3)$, do you get $x^2-3x$?

 

[José Ricardo]

(x - 3)² + (y - 0)² = 0

Did I hit now?

 

[José Ricardo]

If you multiply out $(x-3)(x-3)$, do you get $x^2-3x$?

Ahhh... Thanks, Nich!

 

[José Ricardo]

That can't be correct. A circle has a strictly positive radius.

Moreover, $(x-3)^2 + (y-0)^2 = x^2 + 9 - 6x + y^2$ is not what you were given.

My professor said that the equation doesn't need to do like that in the second member. So he is wrong? [thinking]

 

[José Ricardo]

Folks,

I'm going to the college right now. I'll give you the solution in the night or even in the college if I access the computers from there.

 

[member 587159]

My professor said that the equation doesn't need to do like that in the second member. So he is wrong? [thinking]

I don't know exactly what you mean, but I just meant that if you expand whar you wrote down, you don't get what you started with, so it couldn't be correct.

 

[tnich]

My professor said that the equation doesn't need to do like that in the second member. So he is wrong? [thinking]

Here's a plan. Take $(x-a)^2 +(y-b)^2 = r^2$
and multiply out $(x-a)^2$ and $(x-b)^2$. To get a polynomial in x and y: $x^2 + sx + t + y^2 + uy + v = r^2$. Then set the coefficient of $x$ ($u =$ ?) in this equation with coefficient of $x$ in your circle equation and solve for $a$. Then figure out what $r$ needs to be to make the constant term ($t + v - r^2$) equal to zero as it is in your circle equation.
Edit: In line with @Math_QED's comment, also make sure that all of the other coefficients match between the two equations.

 

[member 587159]

Here's a plan. Take $(x-a)^2 +(y-b)^2 = r^2$
and multiply out $(x-a)^2$ and $(x-b)^2$. To get a polynomial in x and y: $x^2 + sx + t + y^2 + uy + v = r^2$. Then set the coefficient of $x$ ($u =$ ?) in this equation with coefficient of $x$ in your circle equation and solve for $a$. Then figure out what $r$ needs to be to make the constant term ($t + v - r^2$) equal to zero as it is in your circle equation.

I wouldn't recommend this approach, as initially it isn't given that the curve is a circle.

 

[tnich]

I wouldn't recommend this approach, as initially it isn't given that the curve is a circle.

True. On the other hand, if it gives a value for the center and gives a positive value for $r^2$, then it proves that the curve is a circle.

 

[Mark44]

That can't be correct. A circle has a strictly positive radius.

Or a zero radius, for a so-called degenerate circle that consists of a single point. We can also have degenerate ellipses and degenerate hyperbolas, the latter of which consist of two lines.

 

[LCKurtz]

Homework Statement

Identify the following figures in the plane, saying if they are circles, ellipses, hyperboles or parabolas and giving their centers, rays, foci, and guidelines as the case may be:

Homework Equations

x² + y² - 3x = 0

The Attempt at a Solution

x² + y² - 3x = 0

In post #3 Math_QED gave the hint to complete the square with respect to $y$. I think he meant with respect to $x$. That should be your first step, and not doing that is what everyone is hinting at is causing your mistakes.

 

[José Ricardo]

Circle: ~x² + ~y² = R²
(x² - 3)² x= x² - 6x +9
~y = y; x: x - 3, c = (3,0)
(x - 3)² + y² → x = 3
I think now it's right!

 

[symbolipoint]

Homework Statement

Identify the following figures in the plane, saying if they are circles, ellipses, hyperboles or parabolas and giving their centers, rays, foci, and guidelines as the case may be:

Homework Equations

x² + y² - 3x = 0

The Attempt at a Solution

x² + y² - 3x = 0
x(x - 3) + y² = 0
c = (3,0)
(x-3)² + y² = 0
(3-3)² + y² -3 = 0
y = 0
x² + 0² - 3x = 0
x(x-3) = 0
x = 3
y = 0
It's a circle, the two square letters are round perfectly.

I would like you to know my errors.

CIRCLE; because both x and y are to exponent 2, possibly including one or both of the variables to exponent 1. AND coefficient on the exponent-two variables is 1 each. Nothing more is needed for identifying the conic section shape.

 

[José Ricardo]

CIRCLE; because both x and y are to exponent 2, possibly including one or both of the variables to exponent 1. AND coefficient on the exponent-two variables is 1 each. Nothing more is needed for identifying the conic section shape.

Symboli,

I restated, am I right now? It seems that gave the same result. Hm...

 

[symbolipoint]

(x - 3)² + (y - 0)² = 0

CIRCLE; because both x and y are to exponent 2, possibly including one or both of the variables to exponent 1. AND coefficient on the exponent-two variables is 1 each. Nothing more is needed for identifying the conic section shape.

Okay, I was wrong. It might be a degenerate circle.

x^2-3x+y^2=0
-
x^2-3x+(3/2)^2+y^2=(3/2)^2
(x-3/2)^2+y^2=(3/2)^2

So I was correct the FIRST time. The shape IS a circle.
Center is ( 3/2, 0 ) and the radius is 3/2.

 

[José Ricardo]

Okay, I was wrong. It might be a degenerate circle.

x^2-3x+y^2=0
-
x^2-3x+(3/2)^2+y^2=(3/2)^2
(x-3/2)^2+y^2=(3/2)^2

So I was correct the FIRST time. The shape IS a circle.
Center is ( 3/2, 0 ) and the radius is 3/2.

@symbolipoint,

Why 3/2 and not 3? I didn't understand.

 

[symbolipoint]

@symbolipoint,

Why 3/2 and not 3? I didn't understand.

I added the term to "complete the square" to both sides of the equation. Search for your lessons about how to and why, to complete-the-square. The method allows for finding a factorization within an equation to obtain (x-a)^2, and / or (y-b)^2. Too many lessons about this are already available, in your textbook/s, in some YouTube videos, and other places online findable through internet online searches. Tell me if you need reference made to one or two of them.

 

[José Ricardo]

I added the term to "complete the square" to both sides of the equation. Search for your lessons about how to and why, to complete-the-square. The method allows for finding a factorization within an equation to obtain (x-a)^2, and / or (y-b)^2. Too many lessons about this are already available, in your textbook/s, in some YouTube videos, and other places online findable through internet online searches. Tell me if you need reference made to one or two of them.

Ahhhh, now I understood, factoration... And yes, I want video lessons from Youtube. Do you know any video that explanate conics doing look-a-like exercises like that in this topic that I posted?

 

[symbolipoint]

Ahhhh, now I understood, factoration... And yes, I want video lessons from Youtube. Do you know any video that explanate conics doing look-a-like exercises like that in this topic that I posted?

Done! See your private messages.

 

[qspeechc]

(x-a)² = x²-2ax+a²
Right?

So you want to make x² - 3x of the above form, and it is missing the constant term on the end, you need to add that to both sides.

(In case anyone is still interested in this problem.)

 

[José Ricardo]

(x-a)² = x²-2ax+a²
Right?

So you want to make x² - 3x of the above form, and it is missing the constant term on the end, you need to add that to both sides.

(In case anyone is still interested in this problem.)

Thanks a lot, Vertigo. I already solved it.

 

[qspeechc]

Yes, I know, I wrote that for the benefit of anyone else who may be interested in this problem, because it's a standard problem with the standard solution I wrote above. I only answered in this thread after reading all the huffing and puffing (no offence, happens to all of us).