Rectangular/Right hyperbolas through three points

Question

Find all equations of regular hyperbolas passing through the points $A(\alpha,0),B(\beta,0),$ and $C(0,\gamma)$.

Attempt: I am assuming here that regular means right. We can then write the equation of such as $(x-x_0)^2-(y-y_0)^2 =a^2$. Thus by the given we have $(\alpha-x_0)^2 -y_0^2 = a^2, (\beta-x_0)^2 -y_0^2 = a^2,$ and $x_0^2 - (\gamma-y_0)^2 = a^2.$ Therefore $\alpha-x_0 = \pm(\beta-x_0) \implies \alpha = \beta$ or $\alpha - 2x_0 = -\beta$.

I am not sure what to do next because we have a lot of variables and should I break this up into cases then with $x = \beta$ or $\alpha -2x_0 = -\beta$?

Answer 1

A conic section is characterized by five real degrees of freedom. Knowing three distinct points on it will account for three degrees of freedom (not six since each point might still move along the conic without changing it). Assuming that it's a right hyperbola should account for another degree of freedom, so we should end up with a one-parameter family of conics. But how to find them?

One useful tool is projective geometry, where you can use an homogeneous coordinate vector like $[x:y:0]$ to describe a point at infinity. This is also the direction of the asymptotes, since the asymptotes are just tangents at these points at infinity. If $[x_1:y_1:0]$ describes one tangent direction, $[y_1:-x_1:0]$ is the perpendicular one. So together with the given three points, that's five points defining a conic section. You just have to compute its equation.

This $x_1$ and $y_1$ may look like two parameters, but since these are homogeneous coordinates, the only thing that matters is their ratio. You could for example write $x_1=\cos(t),y_1=\sin(t)$ and characterize things in terms of an angle $t$. Adding a multiple of $90°$ to that angle would result in the same conic, since it would be defined by the same set of points. So you could also w.l.o.g. take $y_1=1$ and $x_1\in[-1,1)$ to obtain a unique description of each hyperbola.

I'm using Sage to work out the formula since I'm lazy:

PR.<x1,y1,alpha,beta,gamma> = QQ[]
A = vector((alpha, 0, 1))
B = vector((beta, 0, 1))
C = vector((0, gamma, 1))
D = vector((x1, y1, 0))
E = vector((y1, -x1, 0))
M1 = A.cross_product(B).column()*C.cross_product(D).row()
M2 = A.cross_product(C).column()*B.cross_product(D).row()
M = (E*M1*E)*M2 - (E*M2*E)*M1
M += M.transpose()
M = M.parent()(M/gcd(M.list()))
M

$$\begin{pmatrix} 2 x_{1} y_{1} \gamma & - x_{1}^{2} \gamma + y_{1}^{2} \gamma & - x_{1} y_{1} \alpha \gamma - x_{1} y_{1} \beta \gamma \\ - x_{1}^{2} \gamma + y_{1}^{2} \gamma & -2 x_{1} y_{1} \gamma & - x_{1} y_{1} \alpha \beta + x_{1} y_{1} \gamma^{2} \\ - x_{1} y_{1} \alpha \gamma - x_{1} y_{1} \beta \gamma & - x_{1} y_{1} \alpha \beta + x_{1} y_{1} \gamma^{2} & 2 x_{1} y_{1} \alpha \beta \gamma \end{pmatrix}$$

If you wanted to, you could turn this matrix representation into a polynomial equation, and also plug in $y_1=1$ as discussed above:

$$x_1\gamma\,(x^2-y^2) -(x_1^2-1)\gamma\,xy -x_1(\alpha+\beta)\gamma\,x -x_1(\alpha\beta-\gamma^2)\,y +x_1\alpha\beta\gamma = 0 \quad\text{with }x_1\in[-1,1) $$

I am assuming here that regular means right.

So am I. Googling for “regular hyperbola” I found one match which claims

For a regular hyperbola the eccentricity is $\sqrt{2}$.

According to Wikipedia, that's just the eccentricity of a rectangular i.e. right hyperbola. So I think your interpretation of the term is likely correct.

Answer 2

Use the equation $$ A x^2 + B y^2 + D x + E y = 1 $$

With the points given you have three equations

$$ \begin{align} A \alpha^2 + D \alpha & = 1 \\ A \beta^2 + D \beta & = 1 \\ B \gamma^2 + E \gamma 7 & = 1 \end{align} $$

I include a skew parameter $\zeta = -B/A$ also to parametrize all the curves and together I get the solution

$$\begin{align} A & = -\frac{1}{\alpha \beta} \\ B & = \frac{\zeta}{\alpha \beta} \\ D & = \frac{\alpha+\beta}{\alpha \beta} \\ E & = \frac{\alpha \beta - \zeta \gamma^2}{\alpha \beta \gamma} \end{align} $$

These coefficients give all the curves passing through the points, for every $\zeta \neq 0$.

Note that when $\zeta<0$ the result is an ellipse with a circle at $\zeta =- 1$.

Answer 3

We'll assume, as discussed elsewhere, that the conic in question is a rectangular hyperbola (also referred to in the OP as a right hyperbola).

Suppose a conic passes through the points $A(a,0),B(b,0),C(0,c),D(0,d)$.

Reducing the equation by making $y=0$ or $x=0$, we deduce (see Salmon, Conic Sections, pg 153, Ex 1 for details) that the equation is

$$ c d x^2+2hxy+ab y^2 -c d(a+b)x -ab(c+d)y +a b c d =0, $$ where $h$ is undetermined.

It's a well known result (see Hatton, Projective Geometry, pg 243) that if a rectangular hyperbola passes through the vertices of a triangle it also passes through the orthocentre. It's also easy to see that the orthocenter of $\triangle{ABC}$ is on the $y$-axis, and this determines the $d$ in $D(0,d)$.

By varying $h$ we can run through the family of rectangular hyperbolae passing through $A,B,C$.

For the equation of a rectangular hyperbola, if the coefficient of $x^2$ is $a$, the coefficient of $y^2$ is $-a$. So $$ ab=-cd \implies d=-\frac{ab}{c}, $$ sparing us the trouble of calculating the orthocenter of $\triangle{ABC}$.