I encountered a curve formula in this form:
$y = Ax + \sqrt{ (x - B)^2 + (Ax + C)^2}$
The author who raised it said it was a hyperbola and through online tool I drew its on the screen and confirmed that.
However I don't know how to analysis the curve like what's its asymptote formula?
Or can some turn into a familiar form for me like
$(\frac{x}{a})^2- (\frac{y}{b})^2 = 1$
I do know it is not simple like this one as the curve must be rotated and translated, I just mean that turning it into a standard form like this one plus some factors.
Squaring on both sides you get $$ y^{2}-2Axy-x^{2}+2\left(B-AC\right)x-\left(B^{2}+C^{2}\right)=0 \\ \ \\ y^2-x^2+2axy+2bx+c=0 $$
Now to analyze we can use some characteristic properties of conics:
Writing down the pair of tangents equation $S_1^2=SS_{11}$ $$ (a^{2}(-w^{2})+2bw+c-w^{2})y^{2}+(-a^{2}z^{2}-2abz-b^{2}-c-z^{2})x^{2}+2(a^{2}wz+abw+ac-bz+wz)xy+...=0 $$
Since this is analytically a circle the coefficient of $x^2$ and $y^2$ are equal and the coefficient of xy is 0
This gives us two hyperbola equations which intersect at the focii $$ a^{2}xy+abx+ac-by+xy=0 \\ \ \\ -a^{2}x^{2}+2bx+c-x^{2}=-a^{2}y^{2}-2aby-b^{2}-c-y^{2} $$
Mathematica gives rather unruly expressions as solutions but ig graphing is better to see it
The pair of asymptotes should be easier
We write the equation in standard form with a variable constant and equate the determinant to zero
$$ y^2-x^2+2axy+2bx+c=0 \\ \ \\ \Delta = \begin{vmatrix}a & h &g\\h & b & f\\g&f&c\end{vmatrix} = \begin{vmatrix}-1 & a &b\\a & 1 & 0\\b&0&c\end{vmatrix}=0 \\ \ \\ c=\frac{-b^2}{1+a^2} $$
So the pair of asymptotes are $$ y^{2}-x^{2}+2axy+2bx-\frac{b^{2}}{1+a^{2}}=0 $$