Quadratic equation of two variables

[LagrangeEuler]

Quadratic equation
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
is
(a) elipse when $B^2-4AC<0$
(b) parabola when $B^2-4AC=0$
(c) hyperbola when $B^2-4AC>0$
I found this in Thomas Calculus. However for some values of parameters $A=17$, $C=8$, $B=\sqrt{4 \cdot 17 \cdot 8}$, $D=E=0$, $F=20$ I got just straight line. Where is mistake?

 

[member659127]

There is no mistake. That quadratic equation is the general equation of a conic section. (When you intersect a cone with a plane.) So in special cases you can get a line, intersecting two lines or even a single point. A circle is also possible. Just google "conic section", you'll understand what I mean...

 

[LagrangeEuler]

There is no mistake. That quadratic equation is the general equation of a conic section. (When you intersect a cone with a plane.) So in special cases you can get a line, intersecting two lines or even a single point. A circle is also possible. Just google "conic section", you'll understand what I mean...

Well look my example. In my post for $A=17$, $C=8$, $B=\sqrt{4 \cdot 17 \cdot 8}$, $D=E=0$, $F=20$ I got just straight line. And for that is $B^2-4AC=0$, and it is not parabola.

 

[BvU]

Look at the first wikipedia picture:

and imagine shifting the purple plane to the right until it contains the origin. The 'conic section' is a line in that case

 

[member659127]

Let me expand Thomas' classification then:

(a) elipse (or circle; or point) when B²−4AC<0
(b) parabola (or line) when B²−4AC=0
(c) hyperbola (or two intersecting lines) when B²−4AC>0

the cases in the paranthesis are called degenerate conic sections. (not sure this nomencleture applies to circle though)

 

[LagrangeEuler]

Here is curve for $A=17$, $C=8$, $B=\sqrt{4 \cdot 17 \cdot 8}$, $F=20$. These are two parallel lines.

 

[BvU]

That's funny -- you sure ? With that choice of parameters the first three terms form a square and those are usually non-negative, not -20

 

[LagrangeEuler]

That's funny -- you sure ? With that choice of parameters the first three terms form a square and those are usually non-negative, not -20

Sorry. My mistake. $F=-20$. Other parameters, are the same. So equation is
$$17x^2+\sqrt{4 \cdot 17 \cdot 8}xy+8y^2=20$$

 

[BvU]

Sort of like $x^2 + 2xy +y^2 = 1$ if I try to keep it simple.
The intersection of $z^2=(x+y)^2$ with $z^2=1$ in homogenous coordinates with a determinant 0, leading to one of the degenerate cases

I grant you I have a hard time forming a picture like in #4 ... ( @erbahar ? )

 

[member659127]

Sort of like $x^2 + 2xy +y^2 = 1$ if I try to keep it simple.
The intersection of $z^2=(x+y)^2$ with $z^2=1$ in homogenous coordinates with a determinant 0, leading to one of the degenerate cases

I grant you I have a hard time forming a picture like in #4 ... ( @erbahar ? )

Didn't quite understand the question (I feel like I might have missed the context, sorry)
My initial reply #2 was to give a geometrical answer to the question.

Of course you can always write something like:
$(ax + by + c)^2 = -4$

which is a perfetly legitimate quadratic, however it does have no solution in the real plane. Parallel line example is also of that sort.

eg. $(ax + by + c)(ax + by + d) = 0$

 

[BvU]

OP corrected the value of F so the form is $(ax+by)^2 =$ positive, with two parallel lines. I wondered what picture to form of a plane intersecting a cone in that specific case ...

 

[member659127]

OP corrected the value of F so the form is $(ax+by)^2 =$ positive, with two parallel lines. I wondered what picture to form of a plane intersecting a cone in that specific case ...

Oh, I see now. I don't think it's possible to picture that as an intersection.