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

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:

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

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

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

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

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,

R=QQ[c]
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

$$2x^2+2y^2-3cx=0$$

• 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. Commented Sep 17, 2022 at 8:21
• @jacubero: See the edit. Commented Sep 17, 2022 at 8:38
• How do you determine s in terms of t and obtain $\frac{2}{t^3}x -\frac{3}{2}\frac{\sqrt[3]{4}}{t}= y$? Commented Sep 18, 2022 at 16:49
• @jacubero $-s^3/2=2/t^3$ gives $s=-\sqrt[3]{4}/t.$ 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 $$\displaystyle \frac{3}{4}\sqrt[3]{2}$$.

• Thanks for this correction , I am going to delete my solution. Commented Sep 17, 2022 at 15:06