1. ## Identifying a conic.

Here is a question that I believe that I might have worked out partially, but I have one small sticking point that is really throwing me.

The question:
Identify the following conic. That is, is it a circle, parabola, hyperbola, or ellipse?
Show why.
Equation:
$x^2-4y^2-4x-24y=48$

$x^2-4y^2-4x-24y=48$ can be identified as a hyperbola by executing the following steps:
Rearrange the terms:
$$(x^2-4c+a)+(-4y^2-24y+b)~48-a+b$$
Complete the square:
(this is where my problem lies, I'm not quite sure that my result is correct.)
(x^2-4c+1)-4(y^2+6t+4)=48+1-16

I realize that my answer is incomplete, but I can't quite figure out where I'm going wrong.

2. In order to complete the square:
you need a=4 and b=-36

3. Originally Posted by quikwerk
Here is a question that I believe that I might have worked out partially, but I have one small sticking point that is really throwing me.

The question:
Identify the following conic. That is, is it a circle, parabola, hyperbola, or ellipse?
Show why.
Equation:
$x^2-4y^2-4x-24y=48$

$x^2-4y^2-4x-24y=48$ can be identified as a hyperbola by executing the following steps:
Rearrange the terms:
$$(x^2-4c+a)+(-4y^2-24y+b)~48-a+b$$
Complete the square:
(this is where my problem lies, I'm not quite sure that my result is correct.)
(x^2-4c+1)-4(y^2+6t+4)=48+1-16
You mean "4x", not "4c" but the problem appears to be that you do not know how to "complete the square". Why did you add "1" and "4"?
A "perfect square" is of the form $(x- a)^2= x^2- 2ax+ a^2$. Compare that to $x^2- 4x+ ?$. Where you have "4x", the general formula has "2ax". Okay, for those to be the same, you must have 4= 2a. What is "a"? What is " $a^2$"? That's what you want to add to both sides of the equation.

You have the same problem with the y term. $-4y^2- 24y= -4(y^2+ 6y)$ (Not 6t. You seem awfully careless with what you write!)
Now compare $(y+ a)^2= y^2+ 2ay+ a^2$ and $y^2+ 6y$. You must have 2a= 6 so what is a? What is $a^2$? That is what you want to add, not "4".

I realize that my answer is incomplete, but I can't quite figure out where I'm going wrong.

4. Originally Posted by HallsofIvy
You mean "4x", not "4c" but the problem appears to be that you do not know how to "complete the square". Why did you add "1" and "4"?
A "perfect square" is of the form $(x- a)^2= x^2- 2ax+ a^2$. Compare that to $x^2- 4x+ ?$. Where you have "4x", the general formula has "2ax". Okay, for those to be the same, you must have 4= 2a. What is "a"? What is " $a^2$"? That's what you want to add to both sides of the equation.

You have the same problem with the y term. $-4y^2- 24y= -4(y^2+ 6y)$ (Not 6t. You seem awfully careless with what you write!)
Now compare $(y+ a)^2= y^2+ 2ay+ a^2$ and $y^2+ 6y$. You must have 2a= 6 so what is a? What is $a^2$? That is what you want to add, not "4".
I apologize for all the typos, the internet at my house is down at the moment, so I am using my smartphone to post thus the keyboard spaces out on me and puts a "c" where I mean "x"....
Thanks for your help, though! I'm going to work it out again with your advice and report back with my results.

5. Hello, quikwerk!

Identify the following conic: circle, parabola, hyperbola, or ellipse.
Show why.

. . . $x^2-4y^2-4x-24y\:=\:48$

If it asks us to "Identify" (only), and not about the center or axes,
. . the answer is hyperbola.

There is an "eyeball" approach to identifying conics.

One general form is: . $Ax^2 + By^2 + Cx + Dy + E \:=\:0$

. . If $A = 0$ or $B = 0$: parabola.

. . If $A = B$: circle.

. . If $\,A$ and $\,B$ have the same sign and $A \ne B$: ellipse.

. . If $AB < 0$ ( $\,A$ and $\,B$ have opposite signs): hyperbola.

You may wish to complete-the-square, etc. to determine if we have:

. . a degenerate conic, such as: . $(x-2)^2 + (y-3)^2 \:=\:0$

. . or an "imaginary" conic, such as: . $\dfrac{x^2}{9} + \dfrac{y^2}{4} \:=\:-1$