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).