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.

Question

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 $P = ( 5 \cos R , 3 \sin R)$.

Answer 1

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 $X x+Y y+1=0.$