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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.

enter image description here

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.

SZYoo
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Dec 13 '23 at 17:30
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    Note: Shape of the curve $ax^2+2bxy+cy^2+2dx+2ey+f=0$ is parabola for $\delta = 0$ and $\Delta \neq 0$, where $\Delta=\begin{vmatrix} a & b & d\ b & c & e\ d & e & f \end{vmatrix}$ and $\delta=\begin{vmatrix} a & b\ b & c \end{vmatrix}$. – Anton Vrdoljak Dec 13 '23 at 18:09

1 Answers1

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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:

  1. The pair of tangents drawn from the focus analytically form a circle (though this circle is obviously imaginary)
  2. The equation of asymptotes differ from the equation of hyperbola only by the constant term

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 $$ y^{2}-x^{2}+2axy+2bx-\frac{b^{2}}{1+a^{2}}=0 $$

RandomGuy
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