Identifying a conic.

**MathBlaster47** · February 22nd 2010, 01:15 PM

I have a question that I have worked out for the most part, but there is one small component of the answer that is confusing me.

The question:
Identify the following conic. That is, is it a circle, parabola, hyperbola or ellipse? Show why.

$x^2-4y^2-4x-24y=48$

Working: $<br /> x^2-4y^2-4x-24y=48 \rightarrow (x^2-4x)-(4y^2-24y)=48$
Completing the square for x and y nets $(\frac{-4}{2})^2=+4$ and $(\frac{-6}{2})^2=+9$ ,respectively.
Therefore, $<br /> x^2-4y^2-4x-24y=48 \rightarrow (x^2-4x)-(4y^2-24y)=48\rightarrow(x^2-4x+4)-(y^2-6y+9)=61$

My issue lies in factoring my revised equations, I can do it just fine but the answers don't appear to match up, when I use F.O.I.L on $(x-2)^2$ I get (x^2-2x-2x-4); when I do the same operation on $(x-2)(x+2)$ I get: $(x^2-2x-2x-(-4))$ which looks like the right answer to me....however, $(x-2)(x+2)$ isn't a square.
I have the same problem with the y side of the equation as well.
I think I might be doing something wrong or making a mistake along the way but I can't see what it is.
What am I not getting?

**icemanfan** · February 22nd 2010, 01:22 PM

The equation is a hyperbola because the $x^2$ and $y^2$ terms on the same side of the equation have opposite sign. Also, you made some mistakes. Let me show you where:

$x^2 - 4x - 4y^2 - 24y = 48$ is perfect.

$x^2 - 4x + 4 - 4(y^2 + 6y + 9) = 48 + 4 - 36$ is the next step.

$(x - 2)^2 - 4(y + 3)^2 = 16$

$\frac{(x-2)^2}{16} - \frac{(y+3)^2}{4} = 1$

Also, you FOILED $(x-2)(x+2)$ incorrectly. The $2x$ terms "cancel out" and you have a difference of squares: $x^2 - 4$ .

**MathBlaster47** · February 22nd 2010, 01:43 PM

Originally Posted by icemanfan

The equation is a hyperbola because the $x^2$ and $y^2$ terms on the same side of the equation have opposite sign. Also, you made some mistakes. Let me show you where:

$x^2 - 4x - 4y^2 - 24y = 48$ is perfect.

$x^2 - 4x + 4 - 4(y^2 + 6y + 9) = 48 + 4 - 36$ is the next step.

$(x - 2)^2 - 4(y + 3)^2 = 16$

$\frac{(x-2)^2}{16} - \frac{(y+3)^2}{4} = 1$

Also, you FOILED $(x-2)(x+2)$ incorrectly. The $2x$ terms "cancel out" and you have a difference of squares: $x^2 - 4$ .

F.O.I.L has always been a sticking point for me, but I do understand why 2x cancels out.
For step 2:
$x^2 - 4x + 4 - 4(y^2 + 6y + 9) = 48 + 4 - 36$
You're saying to factor the 4 out of $(4y^2-24y)$ and make it into $4(y^2-6y)$ then complete the square for it.
I understand that, but wouldn't $(x-2)^2$ have the result $(x^2-4y-4)$ as opposed to $(x^2-4x+4)$ ?

**icemanfan** · February 22nd 2010, 01:49 PM

Originally Posted by MathBlaster47

F.O.I.L has always been a sticking point for me, but I do understand why 2x cancels out.
For step 2:
$x^2 - 4x + 4 - 4(y^2 + 6y + 9) = 48 + 4 - 36$
You're saying to factor the 4 out of $(4y^2-24y)$ and make it into $4(y^2-6y)$ then complete the square for it.
I understand that, but wouldn't $(x-2)^2$ have the result $(x^2-4y-4)$ as opposed to $(x^2-4x+4)$ ?

$(x-2)^2 = (x-2)(x-2) = x^2 - 2x - 2x + (-2) \cdot (-2) =$

$x^2 - 4x + 4$ .

Two negatives multiplied together equals a positive.

**MathBlaster47** · February 22nd 2010, 02:05 PM

Originally Posted by icemanfan

$(x-2)^2 = (x-2)(x-2) = x^2 - 2x - 2x + (-2) \cdot (-2) =$

$x^2 - 4x + 4$ .

Two negatives multiplied together equals a positive.

Thanks....I'm still a little confused because I thought with F.O.I.L I was supposed to subtract everything??

**mathemagister** · February 22nd 2010, 02:56 PM

Originally Posted by MathBlaster47

Thanks....I'm still a little confused because I thought with F.O.I.L I was supposed to subtract everything??

The general rule is:

$(a+b)(a-b)=a^2-b^2$ (because the O.I terms cancel out.)

**StonerPenguin** · February 22nd 2010, 03:04 PM

Originally Posted by MathBlaster47

Thanks....I'm still a little confused because I thought with F.O.I.L I was supposed to subtract everything??

No, you ADD everything. Please see Chapter 4-3 (pages 174-175) of the textbook.
Also here's what I did (plus the people on here can check to see if it's right

)

Identify the following conic. That is, is it a circle, parabola, hyperbola or ellipse? Show why.
$x^2-4y^2-4x-24y=48$
$(x^2-4x+?) + (-4y^2-24y+?) = 48 +?+?$ (Complete the square)
$(x^2 - 4x + 4) - 4(y^2 + 6y + 9) = 48 + 4 - 36$
$(x - 2)^2 - 4(y+3)^2 = 16$
$\frac{(x-2)^2}{16} - \frac{(y+3)^2}{4}=1$
And this is the general form of a... hyperbola! =D (See chapters 9-5 and 9-6 but especially page 432)

**Diagonal** · February 22nd 2010, 05:36 PM

Originally Posted by MathBlaster47

F.O.I.L has always been a sticking point for me

Then might I suggest that you don't use it for the moment, but take that extra step until you do get the pattern and so then can do it more readily in future. In the meantime...

(x - 2)(x - 2) = x(x-2) -2(x - 2) = x^2 - 2x -2x + 4 = x^2 - 4x + 4

(x -2)(x + 2) = x( x + 2) - 2(x + 2) = x^2 + 2x -2x -4 = x^2 - 4

It is unfortunate, but a lot of young people today look at something, but don't really "study" it [i.e. examine under the microscope.] You will find it advantageous in future to memorise the patterns for the perfect square and the difference of squares, to of which are shown here. Do many examples until you see the pattern. They are used often in analysis, and make the work so much easier in the long run. Of course, most pairs of factors encountered are quite different form each other and do require working out in full, but FOIL is an advantage only if it is first studied to avoid sign mixups. By that, I mean you'd learn to miss only the first step in each of the above.

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