A point lies on a dilation of a conic with scale factor $k$ iff its polar is tangent to the dilation of that conic with scale factor $1/k$

Question

While reading coordinate geometry I learned a interesting fact.

If the pole of a straight line wrt a conic of parameter $a$ lies on a similar conic with parameter $a/n$ then the straight line is tangent to the similar conic with parameter $na$

Here conic refers to parabola,ellipse,hyperbola and circles as well.The parameter for circle is radius , for parabola it's latus rectum and for ellipse and hyperbola its length of major axis.By similar I mean that for a circle it has the same centre ,for parabola same vertex and axis and for ellipse and hyperbola same centre and major,minor axis

I am not able to imagine intuitively as to why this happens.Are there any interesting problems that directly uses this and is it valid for all cases?

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Proposed restatement by @Blue. For a conic $C$, let $C_{k}$ be a dilation of $C$ with scale factor $k$. For $C$ a parabola, the center of dilation is the vertex of $C$; otherwise, it is the center of $C$.

Conjecture. $$\text{Point $P$ lies on $C_k$ ($k\neq 0$) iff the polar of $P$ with respect to $C$ is tangent to $C_{1/k}$.}$$

For illustration, the figures below show conics $C$, along with dilations $C_2$ (containing $P$) and $C_{1/2}$ (containing $Q$). In each case, the polar of $P$ with respect to $C$ is tangent to $C_{1/2}$, while the polar of $Q$ is tangent to $C_2$.

Answer 1

It seems that a polar of a point with respect to a conic does not depend on the choice of coordinate system (see, for instance [VT, p. 102]; unfortunately, I failed to find a precise formulation of this fact neither in my (electronic) books nor by googling for English and Russian references), so then it suffices to consider the affine coordinate systems in which the conic is determined by canonical equations. Anyway, in [KK] are given the equations of the polar for a conic determined by a general equation (in [VT] a point determining the polar is assumed to be different from the center of a conic), so at least we prove the claims for convenient coordinate systems. We recall the respective equations.

1) Assume that the conic $C$ is an ellipse or a hyperbola. Put the origin of the coordinate system to its center. According to [KK, 2.4-6] the conic $C$ has an equation $$ax^2+2bxy+cy^2+d=0.$$ So an equation of a conic $C_k$ is $$ax^2+2bxy+cy^2+dk^2=0$$
and an equation of a conic $C_{1/k}$ is $$ax^2+2bxy+cy^2+d/k^2=0.$$

An equation of a polar of a point $(x_1,y_1)$ with respect to the conic $C$ [KK, 2.4-10] is $$axx_1+b(xy_1+yx_1)+cyy_1+d=0$$ and an equation of a straight line tangent to the conic $C_{1/k}$ at its point $(x_2,y_2)$ according to [KK, 2.4-10] is $$axx_2+b(xy_2+yx_2)+cyy_2+d/k^2=0.$$

Now we can prove the equivalence.

Point $P$ lies on $C_k$ ($k\neq 0$) iff the polar of $P$ with respect to $C$ is tangent to $C_{1/k}$.

$\Rightarrow$. Assume that a point $P(x_1,y_1)$ lies on the conic $C_k$. Then a point $P’(x_1/k^2,y_1/k^2)$ lies on the conic $C_{1/k}$. An equation of a straight line tangent to the conic $C_{1/k}$ at the point $P’$ is

$$axx_1/k^2+b(xx_y/k^2+yx_1/k^2)+cyy_1/k^2+d/k^2=0.$$

which is the equation of the polar of $P$ with respect to the conic $C$.

$\Leftarrow$. Assume that the polar of $P(x_1,y_1)$ with respect to the conic $C$ is tangent to the conic $C_{1/k}$. That is there exist a point $(x_2,y_2)\in C_{1/k}$ such that the equations

$$axx_1+b(xy_1+yx_1)+cyy_1+d=0$$ and

$$axx_2+b(xy_2+yx_2)+cyy_2+d/k^2=0$$

define the same straight line. Since the conics do not pass through the origin we have $d\ne 0$. Since $ac-b^2\ne 0$ we obtain $(x_1,y_1)=(x_2k^2,y_2k^2)$, that is the point $P$ lies on the conic $C_k$.

2) Assume that the conic $C$ is a parabola. According to [KK, 2.4-8], we can rotate the coordinate axis and move the origin transforming the equation of a the conic $C$ to a form $$y^2=2px.$$

Then the vertex of the parabola is placed at the origin, so an equation of a conic $C_k$ is $$y^2=2pkx$$ and an equation of a conic $C_{1/k}$ is $$y^2=2(p/k)x.$$

An equation of a polar of a point $(x_1,y_1)$ with respect to the conic $C$ according to [KK, 2.4-10] is $$y_1y+p(x+x_1)=0$$ and an equation of a straight line tangent to the conic $C_{1/k}$ at its point $(x_2,y_2)$ according to [KK, 2.4-10] is (1) $$y_2y+p(x+x_2)/k=0.$$

Now we can prove the equivalence.

Point $P$ lies on $C_k$ ($k\neq 0$) iff the polar of $P$ with respect to $C$ is tangent to $C_{1/k}$.

Indeed, the polar $\ell$ of $P$ with respect to $C$ is tangent to $C_{1/k}$ iff there exists a point $P’(x_2, y_2)$ of the conic $C_{1/k}$ such that equation (1) defines the straight line $\ell$. Since $y_2^2=2(p/k)x_2$, equation (1) becomes $2ky_2y+2xp+ky_2^2=0$. This equation can define a straight line $\ell$ with suitable choice of $y_2$ iff $y_1=ky_2$ and $2px_1=ky_2^2$, that is iff a point $P(x_1,y_1)$ lies on the conic $C_k$.

References

[KK] Granino Korn, Theresa Korn Mathematical Handbook for scientists and engineers, 2nd edition, McGraw Hill, 1968 (Russian translation, Moskow, “Nauka”, 1973).

[VT] A. P. Veselov, E. V. Troitskiy Lectures on analytical geometry, Moskow, Center of applied research at faculty of mechanics and mathematics MGU, 2002 (in Russian).

Answer 2

In particular case of ellipses is it like this ?

They are zoomable with respect the pole. Between any two polar rays we have "curvilinear" similar triangles similarly placed in homothety.

The ellipses are placed in a pencil of rays, only three are shown as above.

You can see two tangent lines in red, where at tangent points, $ \psi = \phi - \theta = 0$ where $\phi$ is slope to x-axis, $\theta$ usual polar angle and $\psi $ radius vector to arc tangent angles which are all equal as labelled.

It requires similarity only as seen, $P$ need not necessarily be at the origin.