Identifying Conic Sections and writing them in Standard Form

[PaySwift]

I missed school and don't know how to do this, Quiz tomorrow! Yikes!! Help please?? :|

Identify the following Conic Sections as either a Parabola, Circle, Ellipse, or Hyperbola. Then, write them in the correct standard form.

1. x² + 2x = -128y - 241 - 16y²

2. -2x + y² = -4y - x² - 4

3. -16 - 2y - 2x = x² - y²

4. x² - 1 = -4y - y²

 

[tkhunny]

Re: Identifying Conic Sections and writing them in Standard

Without an xy-term, these present no difficulty. simply complete the square on every variable with both linear and quadratic terms. The 4th one looks fun.

x^2 - 1 = -4y - y^2

Shuffle a little

x^2 + y^2 + 4y = 1

Completing the square, in a couple of steps.

x^2 + y^2 + 4y + ___ = 1 + ___

4/2 = 2
2^2 = 4

x^2 + y^2 + 4y + 4 = 1 + 4

x^2 + (y + 2)^2 = 5

That is a circle with center at (0,-2) and radius sqrt(5).

You show us the next one.

 

[PaySwift]

Re: Identifying Conic Sections and writing them in Standard

Okay, I think I'm understanding you here. So, in the case of this one:

x² + 2x = -128y - 241 - 16y²

Would I complete the square for both the x² and the -16y² ?

x² + 2x + __ + 16y² + 128y + __ = -241 + __ + __ ? Is that how it's done?

 

[PaySwift]

Re: Identifying Conic Sections and writing them in Standard

Well, none of them are parabolas. I know that for sure.

 

[soroban]

Re: Identifying Conic Sections and writing them in Standard

Hello, PaySwift!

Identify the following Conic Sections as either a Parabola, Circle, Ellipse, or Hyperbola.
Then, write them in the correct standard form.

$\displaystyle (1)\; x^2 + 2x \:=\: -128y - 241 - 16y^2$

$\displaystyle (2)\; -2x + y^2 \:=\: -4y - x^2 - 4$

$\displaystyle (3)\; -16 - 2y - 2x = x^2 - y^2$

$\displaystyle (4)\; x^2 - 1 \:=\: -4y - y^2$

If you just want to identify the conic section, we can "eyeball" the equation.


$\displaystyle \text{Bring all terms to the left side.}$
$\displaystyle \text{Compare the coefficients of }x^2\text{ and }y^2.$

. . $\displaystyle \text{If the coefficients have the same sign and are }\begin{Bmatrix}\text{equal: } & \text{circle} \\ \text{unequal:} & \text{ellipse} \end{Bmatrix}$

. . $\displaystyle \text{If the coefficients have opposite signs: hyperbola.}$

. . $\displaystyle \text{If only one of them is squared: parabola.}$


$\displaystyle \begin{array}{cccccc}(1) & x^2 + 16y^2 + 2x + 128y + 241 &=& 0 & \text{ellipse} \\ \\ (2) & x^2 + y^2 - 2x + 4y + 4 &:=& 0 & \text{circle} \\ \\ (3) & x^2 - y^2 + 2x + 2y + 16 &=& 0 & \text{hyperbola} \\ \\ (4) & x^2+y^2 + 4y - 1 & =& 0 & \text{circle} \end{array}$

 

[tkhunny]

Re: Identifying Conic Sections and writing them in Standard

Okay, I think I'm understanding you here. So, in the case of this one:

x² + 2x = -128y - 241 - 16y²

Would I complete the square for both the x² and the -16y² ?

x² + 2x + __ + 16y² + 128y + __ = -241 + __ + __ ? Is that how it's done?

Just be careful with that leading coefficient that is not unity (1). I'd do this:

x² + 2x + __ + 16(y² + 8y + __) = -241 + __ + 16(__)