How can you tell whether the equation of a non-linear relationship represents a parabola, a hyperbola or a circle?

Question

How can you tell whether the equation of a non-linear relationship represents a parabola, a hyperbola or a circle?

Answer 1

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.

Answer 2

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.

$$\tiny \begin{cases} \det(A)=0\;(\text{degenerated}) & \begin{cases} \text{range }1 & \text{real double line} \\ \text{range }2 & \begin{cases} A_{33}=0 & \text{ parallel lines} \\ A_{33}<0 & 2\text{real secant lines} \\ A_{33} & 2 \text{imaginary lines} \begin{cases} A_{22}>0 & imaginary \\ A_{22}<0 & real \end{cases} \end{cases} \end{cases} \\\det(A)\neq0\;(\text{not denegerated, range }3) & & \begin{cases} A_{33}>0 & elipse \begin{cases} a_{11}\det(A)<0 & real \\ a_{11}\det(A)>0 & imaginary \end{cases} \\ A_{33}=0 & real\;parabola \\ A_{33}<0 & real\;hyperbole \end{cases} \end{cases}$$

Answer 3

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.

Answer 4

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.