# Graphical behaviour of $ax^2+by^2+c=0$ depending upon the signs of $a,b,$ and $c$

When I used a graphing calculator to plot $$ax^2+by^2+c=0$$, I observed the following:

When $$a$$ and $$b$$ are of same signs and $$c$$ is of opposite sign, I get either a circle (when $$a=b$$) or an ellipse (when $$a\ne b$$).

This is expected as the equation $$ax^2+by^2+c=0$$ can also be written as $$ax^2+by^2=-c$$ where $$a,b$$ and $$-c$$ are of same sign. If $$a=b$$, the equation can be written as $$x^2+y^2=-c/a$$ and this clearly represents a circle centred at the origin $$(0,0)$$ with radius $$\sqrt{-c/a}$$.

When $$a$$ and $$b$$ are of opposite signs. Two cases arise:

• $$b$$ and $$c$$ are of same sign but $$a$$ is of opposite sign - Here it represents a hyperbola with $$x$$-axis as its axis.

• $$a$$ and $$c$$ are of same sign but $$b$$ is of opposite sign - Here it represents a hyperbola with $$y$$-axis as its axis.

We can express $$ax^2+by^2+c=0$$ as follows:

$$ax^2+by^2=-c$$

$$-\frac a c x^2 -\frac b c y^2=1$$

When $$b$$ and $$c$$ are of same sign but $$a$$ is of opposite sign compared to $$b$$ and $$c$$, the coefficient of $$x^2$$ is positive whereas that of $$y^2$$ is negative. This can alternatively be represented as $$\frac{x^2}{c}-\frac{y^2}{d}=1$$ where $$c$$ and $$d$$ are arbitrary positive constants. This is similar to the equation of hyperbola with $$x$$ axis as its axis. Similarly, it can be proven for the other case.

When all $$a,b,$$ and $$c$$ are of same signs, there is nothing on the graph (tried all zoom levels).

I am unable to explain this last observation unlike the rest. It would be great if you could tell the reason for observing nothing on the graph, when all the three arbitrary constants are of same signs - positive or negative. Further, kindly tell whether I missed any other interesting cases regarding the behaviour of $$ax^2+by^2+c=0$$.

• Related but not the same : Conics and conics of the form $ax^2+by^2+c=0$ – Guru Vishnu Nov 4 '19 at 13:39
• for all $x,y\in\Bbb R$, $x^2, y^2\ge0$ so if $a,b,c>0$ then $ax^2+by^2+c>0$ and if $a,b,c<0$ then $ax^2+by^2+c<0$ – J. W. Tanner Nov 4 '19 at 13:45
• @J.W.Tanner, Thank you. Did I miss any other graphical property regarding the same? – Guru Vishnu Nov 4 '19 at 13:48
• There should be $8$ possibilities for signs of non-zero $a,b,$ and $c$ – J. W. Tanner Nov 4 '19 at 13:54

When $$a,b,c$$ are of the same sign you can think they are positive, since otherwise you can change it. Therefore you get
$$ax^2+by^2+c=0$$
where $$c>0$$. This means that there are no point $$(x,y)$$ which can satisfy this equation and their set is empty. That's why you get no graph.