Classify the following as the equation of a line, parabola, ellipse, hyperbola, or circle.

What would be the easiest way to do this?!

Here are the equations:

$\displaystyle x^2-20x=-y^2-6y$

and

$\displaystyle 13x^2+13y^2-26x+52y=-78$

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- Sep 1st 2007, 01:07 AMdeathtolife04Classify the following as the equation...
Classify the following as the equation of a line, parabola, ellipse, hyperbola, or circle.

What would be the easiest way to do this?!

Here are the equations:

$\displaystyle x^2-20x=-y^2-6y$

and

$\displaystyle 13x^2+13y^2-26x+52y=-78$ - Sep 1st 2007, 01:20 AMqbkr21Re:
- Sep 1st 2007, 01:25 AMtopsquark
Are there

*any*of these that you can do? If not you need a*serious*review!

You need to put the equations in terms of standard form by completing the square. For the first one:

$\displaystyle x^2-20x=-y^2-6y$

$\displaystyle (x^2 - 20x) + (y^2 + 6y) = 0$

$\displaystyle (x^2 - 20x + 10^2 - 10^2) + (y^2 + 6y + 3^2 - 3^2) = 0$

$\displaystyle (x^2 - 20x + 10^2) + (y^2 + 6y + 3^2) = 10^2 + 3^2$

$\displaystyle (x - 10)^2 + (y + 3)^2 = 109$

You tell me what kind of curve this is.

As far as $\displaystyle 13x^2+13y^2-26x+52y=-78$ is concerned you have a typo. This is not a valid equation for real numbers.

-Dan - Sep 1st 2007, 01:41 AMdeathtolife04
- Sep 1st 2007, 03:41 AMearboth
Hello,

as topsquark told you this equation is unvalid:

$\displaystyle 13(x^2-2x+1)+13(y^2+4y+4)=-78+13+52$ completing the squares and dividing by 13:

$\displaystyle (x-1)^2+(y+2)^2=-1$

The LHS of the equation would describe a circle with the centre at C(1, -2) but the RHS has to be a square. And a square never is negative. So ... see above.

If and only if your equation reads: $\displaystyle 13x^2+13y^2-26x+52y=78$ then you'll get:

$\displaystyle (x-1)^2+(y+2)^2=11$ that means the circle has a radius of $\displaystyle \sqrt{11}$ - Sep 1st 2007, 05:28 AMticbol
I don't know if you have studied yet all those: line, parabola, ellipse, hyperbola, circle. If yes, then you may understand my answer. If not all of them yet, then you will not fully understand my answer.

Line.

It is linear or of degree 1. Meaning, the x or the y have exponents of 1 only.

It is in the form, y = Ax +B. Or, Ax +By +C = 0.

Parabola.

It is quadratic, or of degree 2. One of the variables is of degree 1 only, though.

It is in the expanded form, y = Ax^2 +Bx +C. Or x = Ay^2 +By +C.

Circle.

It is quadratic, or of degree 2. Both variables are in degree 2. The coefficients of the two squared variables are equal or the same. It is the sum of the two squared variables, etc.

In its expanded form, Ax^2 +Ay^2 +Bx +Cy +D = 0

See that Ax^2 + Ay^2 ?

Ellipse.

It is quadratic, or of degree 2. Both variables are in degree 2. The coefficients of the two squared variables are unequal or not the same. It is the sum of the two squared variables, etc.

In its expanded form, Ax^2 +By^2 +Cx +Dy +E = 0

See that Ax^2 + By^2 ?

Hyperbola.

It is quadratic, or of degree 2. Both variables are in degree 2. The coefficients of the two squared variables may be equal or not equal. It is the subtraction of the two squared variables, etc.

In its expanded form, Ax^2 -Ay^2 +Bx +Cy +D = 0

Or, Ax^2 -By^2 +Cx +Dy +E = 0

See those Ax^2 -Ay^2, or Ax^2 -By^2 ?

----Sometimes it has the "xy" term in the expanded form, even if the squared variables as not in there. Meaning, if you see a quadratic equation with an "xy" term, sum or subtraction, the equation is hyperbola.

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So, the two equations in your question here are both circles, although the last one is imaginary because the sum of two squared quantities (after completing the squares) cannot be negative. --------answer. - Sep 1st 2007, 05:51 AMtopsquark
I should add that these are most of the forms you are going to meet: the ones where the axes of symmetry lie parallel to the x or y axes. However not all of the conic sections you will run into will have these kinds of axes of symmetry and they will appear in the general form of:

$\displaystyle Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$

It is the presence of the xy term that will tell you that you have one of these. You'll be taught how to handle these in your class in due course if you are going to run into them.

-Dan - Sep 1st 2007, 11:35 AMdeathtolife04