Find the locus of the intersection point of two perpendicular tangents to the curve $xy^2=1$.

I find out that the tangents to this curve with slope $m$ has this general form:

$y = mx+(-2m)^{\frac{1}{3}}-m \left( -\frac{1}{2m} \right)^{\frac{2}{3}}$

The perpendicular tangent would have this form then:

$y = -\frac{1}{m}x+ \left(\frac{2}{m} \right)^{\frac{1}{3}}+\frac{1}{m} \left( \frac{m}{2} \right)^{\frac{2}{3}}$

I solved geometrically that the locus is the circle: $2x^2+2y^2-3\sqrt[3]{2}x=0$ (you can check it here: desmos.com/calculator/xt2xquh72l), but I couldn't solve it algebraically. I wonder if you could help me in eliminating $m$ in this system:

$\begin{cases} y &= mx+(-2m)^{\frac{1}{3}}-m \left( -\frac{1}{2m} \right)^{\frac{2}{3}} \\ y &= -\frac{1}{m}x+ \left(\frac{2}{m} \right)^{\frac{1}{3}}+\frac{1}{m} \left( \frac{m}{2} \right)^{\frac{2}{3}} \\ \end{cases} $

  • $\begingroup$ Are you sure of your first equation (for the tangents of slope $m$)? I find a different one. $\endgroup$ Commented Sep 16, 2022 at 20:35
  • 1
    $\begingroup$ @AnneBauval Yes, I am sure. You can check it here: desmos.com/calculator/xt2xquh72l $\endgroup$
    – jacubero
    Commented Sep 16, 2022 at 21:02

2 Answers 2


The curve is $x=1/t^2,y=t$ with dual curve (The dual curve lives in the dual plane of lines in the plane and is the set of tangents to the original curve. I use the parametric form $X=\frac{-y'}{xy'-yx'},Y=\frac{x'}{xy'-yx'}$) $X=-t^2/3,Y=-2/(3t)$ so two independent tangents are

$(-t^2/3)x -2y/(3t)+1=0 \\ (-s^2/3)x -2y/(3s)+1=0 $

(this is two independent parameters $s,t$ in the universal line $xX+yY+1=0$ that links the plane and the dual plane.)

using that the slope of the perpendicular to slope $m$ is $-1/m$ two perpendicular tangents are (determine $s$ i.t.o. $t$)

$-\frac12 t^3 x +\frac32 t =y\\ \frac{2}{t^3}x - \frac32\frac{\sqrt[3]{4}}{t} =y$

which intersect in

$x = (3 t^2)/(t^4 - 2^\frac23 t^2 + 2 \cdot 2^\frac13)\\y = (3 (2^\frac23 t^3 - 2 \cdot2^\frac13 t))/(2 (-t^4 + 2^\frac23 t^2 - 2 \cdot2^\frac13))$

which implicitizes to your circle.

Edit (or your solution w/o dual curves also leads to the same conclusion)

For your specific question solve the system for $x,y$ (I used wolframalpha Solve[{y == 2^(1/3) (-m)^(1/3) - ((-m^(-1))^(2/3) m)/2^(2/3) + m x, y == 2^(1/3) (m^(-1))^(1/3) + 1/(2^(2/3) m^(1/3)) - x/m}, {x, y}]), then substitute $m=n^3$ to get

$x=(3\cdot 2^\frac13 n^2)/(2 n^4-2 n^2+2) \\y=(3\cdot 2^\frac13 n^3-3\cdot 2^\frac13 n)/(2 n^4-2 n^2+2)$

set $c=2^\frac13$ and implicitize, in say M2,

S=R[n,x,y,MonomialOrder=>Eliminate 1]
I=ideal(c^3-2,(2*n^4-2*n^2+2)*x-3*c*n^2,(2*n^4-2*n^2+2)*y-3*c*(n^3-n)) -- matrix {{c^3-2, 2*x^2+2*y^2-3*c*x, 2*n^2*y+2*n*x-3*c*n-2*y, n^2*x-n*y-x}}

that is, the locus is


  • $\begingroup$ Great solution. I am not an expert in dual curves. If you could ellaborate how you find out dual curves and tangents, it would be great. $\endgroup$
    – jacubero
    Commented Sep 17, 2022 at 8:21
  • 1
    $\begingroup$ @jacubero: See the edit. $\endgroup$ Commented Sep 17, 2022 at 8:38
  • $\begingroup$ How do you determine s in terms of t and obtain $\frac{2}{t^3}x -\frac{3}{2}\frac{\sqrt[3]{4}}{t}= y$? $\endgroup$
    – jacubero
    Commented Sep 18, 2022 at 16:49
  • 1
    $\begingroup$ @jacubero $-s^3/2=2/t^3$ gives $s=-\sqrt[3]{4}/t.$ $\endgroup$ Commented Sep 18, 2022 at 18:05

The parametric equations for the curve $xy^2=1$ could be $\forall \ t \neq 0$:

$\begin{cases} x =& \displaystyle t^2 \\ y =& \displaystyle \frac{1}{t} \\ \end{cases}$

The tangent at the point $\left( t^2, \frac{1}{t} \right)$ is:

$m = \displaystyle \frac{\mathrm{d} y}{\mathrm{d}x} = \frac{ \frac{\mathrm{d}y}{\mathrm{d}t} }{ \frac{\mathrm{d}x}{\mathrm{d}t} } = \frac{ -\frac{1}{t^2} }{ 2t } = -\frac{1}{2t^3}$

The tangent equation to the curve in the point $\left( t^2, \frac{1}{t} \right)$ passing through $(x,y)$ is then:

$y - \frac{1}{t} = -\frac{1}{2t^3} \left( x-t^2 \right) \implies 2yt^3-3t^2+x=0$

Using Vieta's formulas to the previous cubic equation la anterior in $t$, their roots $t_1$, $t_2$ y $t_3$ satisfy:

$\begin{cases} t_1+t_2+t_3 &= \frac{3}{2y} \\ t_1t_2+t_2t_3+t_1t_3 &= 0 \\ t_1t_2t_3 &= -\frac{x}{2y} \end{cases}$

If we want two of these tangents to be perpendicular, the condition $m_1 m_2=-1$ must be satified, where $m_1$ and $m_2$ are their slopes. Then:

$\begin{array}{ll} & \displaystyle m_1 \cdot m_2 = -1 \implies \left( -\frac{1}{2t_1^3} \right) \left( -\frac{1}{2t_2^3} \right) = \frac{1}{4t_1^3t_2^3}=-1 \implies 4t_1^3t_2^3 =-1 \implies \\ & \displaystyle t_1t_2 = -\frac{1}{\sqrt[3]{4}} \end{array}$

Substituting this value in Vieta's formulas:

$\begin{array}{ll} & \displaystyle \begin{cases} t_1+t_2+t_3 &= \frac{3}{2y} \\ -\frac{1}{\sqrt[3]{4}}+t_2t_3+t_1t_3 &= 0 \\ -\frac{1}{\sqrt[3]{4}} t_3 &= -\frac{x}{2y} \end{cases} \implies \begin{cases} t_1+t_2+\frac{\sqrt[3]{4}x}{2y} &= \frac{3}{2y} \\ -\frac{1}{\sqrt[3]{4}}+\frac{\sqrt[3]{4}x}{2y}t_2+\frac{\sqrt[3]{4}x}{2y}t_1 &= 0 \\ t_3 &= \frac{\sqrt[3]{4}x}{2y} \end{cases} \implies \\ & \displaystyle \begin{cases} t_1+t_2 &= \frac{3- \sqrt[3]{4}x}{2y} \\ \frac{\sqrt[3]{4}x}{2y} \left( t_1 +t_2 \right) &= \frac{1}{\sqrt[3]{4}} \\ \end{cases} \implies \frac{\sqrt[3]{4}x}{2y} \cdot \frac{3- \sqrt[3]{4}x}{2y} = \frac{1}{\sqrt[3]{4}} \implies \\ & \displaystyle 2x^2+2y^2-3 \sqrt[3]{2} x=0 \iff \left( x -\frac{3}{4}\sqrt[3]{2} \right)^2 + y^2 = \left( \frac{3}{4}\sqrt[3]{2} \right)^2 \end{array}$

Consequently, the locus is the circle centered in $\displaystyle \left( \frac{3}{4}\sqrt[3]{2},0 \right)$ and radius $\displaystyle \frac{3}{4}\sqrt[3]{2}$.

  • $\begingroup$ Thanks for this correction , I am going to delete my solution. $\endgroup$
    – Z Ahmed
    Commented Sep 17, 2022 at 15:06

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