Find domain and range of the slanted hyperbola

Question

Given the conic section $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ and I know that it is a hyperbola and $B\ne 0$.

How to find its domain and range? I guess the method of Lagrange multipliers will fail here.

Answer 1

To find a range, fix $y$ and consider a quadratic equation with respect to $x$: $$Ax^2+(By+D)x+C^2y+Ey+F=0$$ If $A=0$ and $B=0$, there is $x=-\frac{C^2+Ey+F}{D}$ for each $y$ and the domain is $(-\infty,\infty)$ ($D!=0$, because, otherwise, the conic is degenerated).

If $A=0$ and $B\ne 0$, there is no $x$ for $y=-\frac{D}{B}$, for other values of $y$ it is $x=-\frac{C^2+Ey+F}{By+D}$ and the domain is $(-\infty,\infty)\setminus \{-\frac{D}{B}\}$.

If $A!=0$, qudratic equation has at least one solution, when its discriminant $d=(By+D)^2-4A(C^2+Ey+F)\ge0$.

So, we need to find $y$, for which $(By+D)^2-4A(C^2+Ey+F)\ge0$.

Rewrite: $$(B^2-4AC)y^2+(2BD-4AE)y+D^2-4AF\ge0$$ If $B^2-4AC=0$, then

1. If $2BD-4AE=0$, the range is $(-\infty,\infty)$ if $D^2-4AF\ge0$ and $\emptyset$ otherwise.
2. The range is $\left[-\frac{D^2-4AF}{2BD-4AE}, \infty\right)$ if $2BD-4AE>0$.
3. The range is $\left(-\infty, -\frac{D^2-4AF}{2BD-4AE}\right]$ if $2BD-4AE<0$.

The discriminant is $d_1=(2BD-4AE)^2-4(B^2-4AC)(D^2-4AF)$.

1. If $d1<0$, the range is $(-\infty,\infty)$ if $B^2-4AC>0$ and $\emptyset$ otherwise.
2. If $d1=0$, the range is $(-\infty,\infty)$ if $B^2-4AC>0$ and $\{-\frac{2BD-4AE}{2(B^2-4AC)}\}$ otherwise.
3. If $d1>0$, there are two roots $y_1=\frac{4AE-2BD-\sqrt{d_1}}{2(B^2-4AC)}$ and $y_2=\frac{4AE-2BD+\sqrt{d_1}}{2(B^2-4AC)}$. The range is $\left(-\infty,y_1\right]\cup\left[y_2,\infty\right)$ if $B^2-4AC>0$ and $[y_1,y_2]$ otherwise.

Same steps are for domain.