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I have the hyperbola $$x^2 + \frac{2}{\sqrt{3}} x y - y^2 = 1$$ and I want to find the foci, but the only resources I can find that talk about finding the foci require the formula to be in standard form, $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

I don't think I can convert my hyperbola to standard form because it isn't purely horizontal or vertical. How do I go about finding the foci in this case?

Blue
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Lawton
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1 Answers1

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Notice first of all that your hyperbola is centred at $O=(0,0)$. If $P=(x,y)$ is a vertex of the hyperbola, then the tangent at $P$ is perpendicular to $PO$, that is $y'=-x/y$. You can compute $y'$ by differentiating the equation of the ellipse: $$ y'={x+y/\sqrt3\over y-x/\sqrt3}. $$ Equating that to $-x/y$ gives: $$ x^2-y^2-2\sqrt3 xy=0, $$ which can be combined with the equation of the hyperbola to obtain $$ P=\left(\pm\sqrt{2\sqrt3+3\over8}, \pm\sqrt{2\sqrt3-3\over8}\right). $$

But yours is a right hyperbola (coefficients of $x^2$ and $y^2$ are opposite), hence we immediately obtain the foci as: $$ F=\sqrt2 P=\left(\pm\sqrt{2\sqrt3+3\over4}, \pm\sqrt{2\sqrt3-3\over4}\right). $$

Intelligenti pauca
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  • I verified that your result is correct, but I don't understand all the steps you took to get there. Could you go into more detail? Specifically, how did you get from $y'$ to $x^2 - y^2 - 2 \sqrt{3} x y = 0$, and from that to $P$? – Lawton Mar 05 '21 at 00:35
  • @Lawton That's trivial algebra, I'm afraid. – Intelligenti pauca Mar 05 '21 at 06:59
  • Taking the idea of the axes above for $ax^2+bxy+cy^2+dx+ey+f=0$ centered in $((2cd-be)/(b^2-4ac)),-(bd-2ae)/(b^2-4ac))$ (so not a parabola):

    $$y'=-(by+2ax+d)/(2cy+bx+e)=-(x-(2cd-be)/(b^2-4ac))/(y+(bd-2ae)/(b^2-4ac))$$ which in the numerator becomes $$(b^3-4abc)x^2+(-2ab^2+8a^2c+2b^2c-8ac^2)xy+(-b^3+4abc)y^2+(-2abd-2bcd+4a^2e+2b^2e-4ace)x+(-2b^2d+4acd-4c^2d+2abe+2bce)y-bd^2+2ade-2cde+be^2=0$$ i.e. a line pair; the axes of a general conic (even when the quadratic terms vanish you get the one line axis of the then parabola).

     – Jan-Magnus Økland Mar 05 '21 at 10:53