A new property of conics?

Question

Prove a property of conics connecting ( variables on arc) focal radius $r$, pedal length /normal $d$ dropped onto tangent from focus and (constants) semi latus-rectum $p$ and eccentricity $e$ :

$$ \frac{2p}{ r}-\frac{p^2}{d^2} = 1- e^2 \tag1$$

which is a new $ (r-\,d) $ relation.

For a parabola special case $$ p= \frac{2 d^2}{r} $$

Are these properties hitherto known ?

The above (1) is intrinsic, based earlier curvature properties I had derived. We can relate to its particular canonical Newton polar form $ (r\,- \theta) , $ when major axis not along $x-$ axis $ \alpha \ne 0$ .

$$ \dfrac{p}r = 1-e \cos(\theta - \alpha) \,\tag2 $$

We obtain a $ (d-\theta) $ relation:

$$ \dfrac{p^2}{d^2} = e^2 -2e \cos (\theta-\alpha) +1\,\tag3 $$

The three equations take two variables at a time from $ (r,\theta, d\,)$ and link them with two constants $ (p,e) $.

John Bentin points out that Equn(1) the Dark Kepler Problem/Petr Blashke Ex 4 conics equation in pedal coordinates from a most recent publication AIP Wiki reference ( June 2017) !

$$ \dfrac{L^2}{d^2} - \dfrac{2 M}{r} = c \tag4 $$

which can be geometrically interpreted to (1) by adjusting physical (gravitational) terms.

Equation (1) is general for a conic.

In Pedal Equn Wiki assigned values $ (-1,0,1)$ give particular cases (ellipse,hyperbola and parabola) respectively to the following expression.

$$ (\frac{b^2}{d^2}-\frac{2a}{r}) \tag5 $$

Answer 1

This is a very nice result. It is un-obvious enough to have been overlooked hitherto—but I am no authority on this! I have sketched my proof below, for the case of an ellipse (the case of a hyperbola may be treated similarly), in confirmation of your claim.

Choosing cartesian axes as the axes of the ellipse, we may write a general point $T$ on an ellipse as $(a\cos\phi,b\sin\phi)$, where $a$ and $b$ are constants with $a\geqslant b$. (The parameter $\phi$ is chosen to avoid confusion with a polar angle $\theta$ or $\theta-\alpha$ in the question.) We have the usual relationships $b^2=a^2(1-e^2)$ and $p=b^2/a$. Using the standard formula for the distance $d$ of a point, here the focus $(ae,0)$, from a line, here the tangent at $T$, namely $bx\cos\phi+ay\sin\phi-ab=0$, we get, after some simplification, $$d^2=\frac{p^2}{1-e^2}\left(\frac2{1+e\cos\phi}-1\right).$$Also, a calculation of $r^2$ using Pythagoras, and more simplification, yields $$\frac pr=\frac{1-e^2}{1-e\cos\phi}.$$ Combining these formulae with further (algebraic) simplification finally gives the result $$\frac{2p}r-\frac{p^2}{d^2}=1-e^2.$$

Added later Some searching has dug up this page for the pedal equation of a conic. So it has to be said that your result isn't new.