# Point of contact between an ellipse $(5 \cos t , 3 \sin t)$ and an apparent tangent $x \cos(R) + y \sin(R) =D$ sliding on it.

Let $$E$$ be an ellipse defined by $$(x^2 / 5^2) + ( y^2 / 3^2) = 1$$ or, equivalently $$( 5\cos t , 3\sin t )$$ with $$0\leq t \leq 2 \pi$$.

Let $$P= ( 5 \cos R , 3 \sin R) \space 0\leq R \leq 2 \pi$$ be a point moving on E.

Let $$D$$ be the distance from $$P$$ to the origin : $$D = \sqrt {(5 \cos R)^2 + (3 \sin R)^2 }$$.

The following straight line , defined by $$x = 5= D$$ when $$R=0$$ , rotates around $$E$$, and, apparently, is always tangent to $$E$$ :

$$x \cos R + y \sin R =D$$.

Desmos construction : https://www.desmos.com/calculator/dmj3tunnoo

My question is : if this straight line is actually tangent to $$E$$ for all values of $$R$$, what is its ( moving) point of contact with $$E$$? Surprisingly ( to me) this point is not $$P = ( 5 \cos R , 3 \sin R)$$.

• Hint: in the parametrization of the ellipse: $x=5\cos R, y=3\sin R$, $R$ does not represent the polar angle of the point (although it does when $R=0,\pi/2$, etc. Commented Sep 25, 2023 at 1:21
• If you draw a circle centered at the origin with radius $D$, your line will be tangent to that circle at the point $(D\cos R,D\sin R)$. Commented Sep 25, 2023 at 3:07
• To find the point of tangency, solve the system of equations $$\frac{x^2}{25} +\frac{y^2}9 = 1\\x\cos R + y \sin R = D$$ for a common solution. If exactly one exists, it will be the point of tangency, as that is the only way a line can intersect an ellipse in a single point. If none exist for some $R$, the line does not intersect the circle. If there are two solutions, the line intersects in two points, and thus is not tangent. Commented Sep 25, 2023 at 22:11
• $(\frac{a^2\cos{R}}{D},\frac{b^2\sin{R}}{D})$ Commented Sep 28, 2023 at 10:01
• @Ian-Magnus. Wonderful, thanks! Commented Sep 28, 2023 at 22:10

The tangency point for your tangent is

$$(\frac{a^2\cos{R}}{D},\frac{b^2\sin{R}}{D})$$

What I actually did was (in maxima CAS)

solve([(x/a)^2+(y/b)^2-1,x*cos(t)+y*sin(t)-sqrt((a*cos(t))^2+(b*sin(t))^2)],[x,y]);


I'm still trying to find an elegant way to see this.

However, the way I was taught to get the tangents was to use the dual curve: parametrize the ellipse: $$(x,y)=(a\frac{1-t^2}{1+t^2},b\frac{2t}{1+t^2})$$ then the dual is $$(X,Y)=(-\frac{y'}{xy'-yx'},\frac{x'}{xy'-yx'})=(-\frac1{a}\frac{1-t^2}{1+t^2},-\frac1{b}\frac{2t}{1+t^2})$$ i.e. $$X x+Y y+1=0$$ is the tangent of the ellipse that has tangency point $$(x,y).$$

Or if you want $$(x,y)=(\sin{R},\cos{R}):$$ $$(X,Y)=(-\frac{y'}{xy'-yx'},\frac{x'}{xy'-yx'})=(-\frac1{a}\cos{R},-\frac1{b}\sin{R}),$$ and again the tangent at $$(x,y)$$ is $$X x+Y y+1=0.$$