1. ## Quadratic in 2 variables (graphs)

Hi,

I have been looking at quadratics of the form:

$ax^2 + bxy + cy^2 + dx + ey + f = 0$

From this I determined that:

$y = \frac{-bx - e \pm \sqrt{(b^2 - 4ac)x^2 + (2be - 4cd)x + (e^2 - 4cf)}}{2c}$

In particular I am looking at xy - 4x - 2y = 0 which is in standard form with:

a = 0, b = 1, c = 0, d = -4, e = -2, f = 0

However the above formula fails since the denominator would be 0. So I approached differently:

xy - 2y = 4x
y(x - 2) = 4x
$y = \frac{4x}{x - 2}$

So I used a graphing program to sketch this:

I am asked to "identify the type of graph obtained" - what sort of graph is this? I would say hyperbola but I don't think that's necessarily right?

2. Originally Posted by r45
Hi,

I have been looking at quadratics of the form:

$ax^2 + bxy + cx^2 + dx + ey + f = 0$

From this I determined that:

$y = \frac{-bx - e \pm \sqrt{(b^2 - 4ac)x^2 + (2be - 4cd)x + (e^2 - 4cf)}}{2c}$

In particular I am looking at xy - 4x - 2y = 0 which is in standard form with:

a = 0, b = 1, c = 0, d = -4, e = -2, f = 0

However the above formula fails since the denominator would be 0. So I approached differently:

xy - 2y = 4x
y(x - 2) = 4x
$y = \frac{4x}{x - 2}$

So I used a graphing program to sketch this:

I am asked to "identify the type of graph obtained" - what sort of graph is this? I would say hyperbola but I don't think that's necessarily right?
you are correct ... it is a hyperbola.

3. Originally Posted by r45
$ax^2 + bxy + cx^2 + dx + ey + f = 0$

From this I determined that:

$y = \frac{-bx - e \pm \sqrt{(b^2 - 4ac)x^2 + (2be - 4cd)x + (e^2 - 4cf)}}{2c}$
Huh? There's no y^2 in original equation

4. Yes, it's a hyperbola. Any quadratic in two variables is a "conic section" (provided you include "degenerate" forms). Since this "looks like" a hyperbola and the only conic section that looks like that is the hyperbola, it is a hyperbola! (Obviously that vertical line between the two branches is an artifact of the graphing program.) Note that xy= 1 has no $x^2$ or $y^2$ term but is a hyperbola.

More specifiically, if you let $x= x'cos(\theta)- y'sin(\theta)$ and $y= x'sin(\theta)+ y' cos(\theta)$ then $xy= x'^2 cos(\theta)sin(\theta)+ x'y'(cos^2(\theta)- sin^2(\theta)- y'sin^2(\theta)$. Now if $\theta= \pi/4$, $sin(\theta)= cos(\theta)= \sqrt{2}/2$ so that $cos^2(\theta)- sin^2(\theta)= 0$ and $xy= x'^2/\sqrt{2}- y'^2/\sqrt(2)$.

Then $xy- 4x- 2y= x'^2/\sqrt{2}- y'^2/\sqrt{2}- 6x'/\sqrt{2}+ 2y'/\sqrt{2}$.

You could now determine the center of the hyperbola by completing the square. Also, a "standard" hyperbola, of the form $\frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$ has axes of symmetry along the x and y axes. Here it has been rotated $\pi/4$ radians or 45 degrees.

About those "degenerate cases": Obviously, the quadratic equation $x^2+ y^2= 0$ is satisfied only by the single point (0, 0) while the set of points satisfying x^2+ y^2= -1 is the empty set. We can think of those as "degenerate" circles. $x^2- y^2= 0$ can be written as (x- y)(x+ y)= 0 so will be satisfied by (x, y) so that either y= x or y= -x, a pair of lines intersecting at the origin. We can think of that as a "degenerate" hyperbola. Finally, $x^2- 2xy+ y^2= 0$ is the same as $(x- y)^2= 0$ whose graph is the single line y= x. We can think of that as a "degenerate" parabola.

5. Originally Posted by r45
Hi,

I have been looking at quadratics of the form:

$ax^2 + bxy + c\textcolor{red}{y^2} + dx + ey + f = 0$
typo

6. Right thanks a lot for the responses.

So in general, if we have:

$y = \frac{ax}{x-b}$

This would represent a "standard" hyperbola except rotated anti-clockwise by $\frac{\pi}{a}$ radians, and translated by b units on the y-axis?

7. Or would that be translated by $b^2$ units?

8. Originally Posted by r45
Right thanks a lot for the responses.

So in general, if we have:

$y = \frac{ax}{x-b}$

This would represent a "standard" hyperbola except rotated anti-clockwise by $\frac{\pi}{a}$ radians, and translated by b units on the y-axis?
Originally Posted by r45
Or would that be translated by $b^2$ units?
Neither. That becomes xy- by= ax or xy- ax- by= 0. That will be a hyperbola rotated through $\pi/4$ radians. The angle of rotation is determined solely by the quadratic terms as I showed before.

It would be translated horizontally by b/2 and vertically by a/2.

9. Ah OK right I just re-read through your earlier post and I understand now.

I also have a follow up question:

If $f(x,y) = ax^2 + bxy + cy^2 + dx + ey + f$ can be factored into a product of degree one factors, what can we say about the graph of f(x,y) = 0?

10. There are a number of self-evident points we could make, like the number of distinct factors would equal the total number of intersections with the x and y axes. But I have a feeling the question is seeking something less obvious?

11. Originally Posted by r45
Ah OK right I just re-read through your earlier post and I understand now.

I also have a follow up question:

If $f(x,y) = ax^2 + bxy + cx^2 + dx + ey + f$ can be factored into a product of degree one factors,
what can we say about the graph of f(x,y) = 0?
WHERE is y^2 ?????????????

12. Ahh sorry I'm hopeless! I've edited it to show the correct term now.

13. Since it is quadratic, saying f(x) can be factored into "a product of degree one factors" means it is either f(x,y)= (ax+ by)(cx+ dy)= 0 or (ax+ by)^2= 0.

In the first case, the graph is the two straight lines ax+ by= 0 and cx+ dy= 0, which I referred to before as a "degenerate hyperbola". In the second case, the graph is the single straight line, ax+ by= 0, which is a "degenerate parabola".

14. So if d, e or f are non-zero then the quadratic $f(x,y) = ax^2 + bxy + cy^2 + dx + ey + f$ cannot be factored into a product of degree one factors?