How can you tell whether the equation of a non-linear relationship represents a parabola, a hyperbola or a circle? How can you tell whether the equation of a non-linear relationship represents a parabola, a hyperbola or a circle?
 A: if you have the equation of $\mathcal{C}$:  $$Ax^2+2B'xy+Cy^2+2D'x+2E'y+F=0$$ You can take $$A=\begin{pmatrix} A & B' & D' \\ B' & C & E' \\ D' & E' & F \end{pmatrix}$$ as their ("characteristic matrix", I dont remember the name, but it is in homogeneous coordinates.)
Then you have to check $\det(A)=I$, If I=0, then $\mathcal{C}$ is degenerated, if $I\neq0$, then $\mathcal{C}$ is not degenerated
let's define $\Delta=B^2-4AC$
$$\begin{cases} \Delta=0 & \text{parabola} \\ \Delta<0 & \text{hyperbola} \\ \Delta>0 & \text{elipse} \end{cases} $$
or their degenerated case.
$$\tiny \begin{cases} \det(A)=0\;(\text{degenerated}) & \begin{cases} \text{range }1 & \text{real double line} \\ \text{range }2 & \begin{cases} A_{33}=0 & \text{ parallel lines} \\ A_{33}<0 & 2\text{real secant lines} \\ A_{33} & 2 \text{imaginary lines} \begin{cases} A_{22}>0 & imaginary \\ A_{22}<0 & real \end{cases} \end{cases} \end{cases}
\\\det(A)\neq0\;(\text{not denegerated, range }3) & & \begin{cases} A_{33}>0 & elipse \begin{cases} a_{11}\det(A)<0 & real \\ a_{11}\det(A)>0 & imaginary \end{cases} \\ A_{33}=0 & real\;parabola \\ A_{33}<0 & real\;hyperbole \end{cases}  \end{cases}$$
A: If you are looking at a relation $$\pm\frac{(x-h)^2}{a^2}\pm\frac{(y-k)^2}{b^2}=1$$
then the tell-tale signs are in the $\pm$, $a$, and $b$.
If $a=b$ and both $\pm$ are $+$, then it is a circle.
If $a\neq b$ and both $\pm$ are $+$, then it is an ellipse.
If only one $\pm$ is $+$, then it is a hyperbola.
These are the only possible cases, since both $\pm$ cannot be $-$ as this would result in a negative number being equal to a positive.
A: Suppose $\dfrac {x^2} {a^2} - \dfrac {y^2} {b^2} = 1$.
If $y=0$, you have $x=\pm a$, so you get two points on the $x$-axis.  If $x=0$, you have $y=\pm\sqrt{-a}$, and since we're only using real numbers here, you get no points on the $x$-axis.  That much alone will tell you something.
Now suppose $x$ is some immense number, say in the billions.  By comparison to that number, the difference between $1$ and $0$ is negligible, so it's as if you had $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2}= 0$, which says $\dfrac x a = \pm \dfrac y b$.  But $\dfrac x a = \dfrac y b$ is a straight line through the origin and $\dfrac x a = -\dfrac y b$ is another straight line through the origin.  That also tells you something.
Now suppose Suppose $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$. If $y=0$ you get $x=\pm a$, and if $x=0$ then $y=\pm b$, so you get two points on the $x$-axis and two points on the $y$-axis, and notice also the symmetry of the whole thing: interchanging $x$ and $-x$ results in the same value of $x^2$ and similarly $y$ and $-y$, so you have symmetry about both axes.  That should also tell you something about what the picture looks like.
A: Let the coefficients of $ x^2, y^2, x y $ be  $ a,b,2 h $ respectively.
Then depending on sign of $ (  a b -h^2 )  i.e.,  ( >0, <0, =0 )$ we have ellipse, hyperbola and parabola respy.
