# Prove that these two definitions of a parabola are equivalent.

$i$: A parabola is the locus of points in the plane equidistant from a fixed point and a fixed line.

$ii$: A parabola is the intersection of a right circular cone with a plane parallel to another plane which is tangent to the surface of the cone.

Prove that these definitions are equivalent. Hint: Reduce the problem to a single case by using the fact that translations, rotations, and reflections preserve distance.

I think the hint is alluding the fact that all parabolas are similar, so should I only consider one parabola (i.e. $y=x^2$) when proving this? Also, how do I go about proving it? I have no idea.

Here is a proof by simply applying some three dimensional Euclidean geometry without using any computations, purely synthetically.

Let $C$ be the circular cone with apex $A$ and axis $a$ (observe $a$ passes through $A$). Let $\omega_C$ be a plane tangent to the cone $C$ (then $A \in \omega_C$) and let $\omega$ be another plane parallel to $\omega_C$ intersecting $C$ in a curve (conic section $\omega \cap C \,$). Then there is a line $l$ on the cone $C$ (generator of $C$ ? It passes through $A$) which lies in $\omega_C$ and is therefore the line where $\omega_C$ is tangent to $C$.

Let $S$ be the sphere tangent to the cone $C$ and to the plane $\omega$ (the center of $S$ is on the axis $a$). First, $S$ is tangent to the cone $C$ in a whole circle called $k$. Second, $S$ is tangent to the plane $\omega$ at a point $F$, the future focus of the parabola.

Denote by $\sigma$ the plane defined by the circle $k$ and let $O \in \sigma$ be the center of $k$. Notice that $O$ also lies on the axis $a$, so $O = a \cap \sigma$. Let line $d$, the future directrix of the parabola, be the intersection line of planes $\sigma$ and $\omega$ and let $t_k$ be the intersection line of $\sigma$ and $\omega_C$. Since $\omega$ and $\omega_C$ are parallel, then the lines $d$ and $t_k$ are parallel. Notice that since $a$, as an axic of the cone, is orthogonal to $\sigma$, it is orthogonal to $t_k$. Moreover, $t_k$ is tangent to circle $k$ at point $T_S$, to orthogonal to the radius $OT_S$ of the circle. Consequently, $t_C$ is orthogonal to the plane $\alpha$ formed by $a$ and $OT_S$ (observe that then $l$ is contained in $\alpha$), so $t_C$ and thus $d$, as parallel to $t_C$ are both orthogonal to $l$. Let $l_{\omega}$ be the intersection line of plane $\alpha$ and plane $\omega$ (then $l_{\omega}$ is the future axis of the parabola). Then $l_{omega}$ is parallel to $l$ and is therefore orthogonal to $d$ (the future directrix).

Pick an arbitrary point from the intersection locus of plane $\omega$ and the cone $C$. Let $X_d$ be the orthogonal projection of $X$ on the line $d$ (both are in plane $\omega$) and let $X_S$ be the intersection point of the line $XA$ (line $XA$ lies on the cone $C$ ) with the circle $k \subset S$. Then as $XA$ is on the cone, it is tangent to the sphere $S$ and its point of tangency is $X_S$. However, since $S$ is tangent to plane $\omega$ at point $F$, then $XF$ is also tangent to the sphere $S$. Two tangents $XX_S$ and $XF$ to the same sphere $S$ from the same point $X$ outside the sphere, have equal lengths, i.e. $XX_S = XF$.

Define the plane $\alpha_X$ to be the plane determined by the two parallel lines $l$ and $XX_d$. Then by construction plane $\alpha_X$ intersects plane $\sigma$ in the line $X_dT_S$ (recall that $T_S \in l$ is the point of tangency of line $t_k$ and circle $k$ lying in plane $\sigma$). But line $X_dT_S$ lies in $\alpha_X$ and intersects line $AX$. However, the only intersection point of $AX$ and plane $\sigma$ containing line $x_dT_S$ is $X_S \in k$, so point $X_S$ lies on $X_dT_S$. Now, since $AX_S$ and $AT_S$ are tangents to the sphere $S$ from point $A$, they are equal, i.e. $AX_S = AT_S$ which means that triangle $X_SAT_S$ is isosceles. But in plane $\alpha_X$ the lines $AT_S \equiv l$ and $XX_d$ are parallel so triangles $X_SAT_S$ and $X_dXX_S$ are similar, yielding that triangle $X_dXX_S$ is also isosceles, meaning that $XX_d = XX_S$. But we have already proven that $XX_S=XF$, so finally:

In the plane $\omega$, where the intersection locus $\omega \cap C$ of the plane $\omega$ with the cone $C$ lies, we have found a line $d$ and a point $F$ with the property that for any point $X$ from the locus $\omega \cap C$ we have that $\text{dist}(X,d) = XX_d = XF = \text{dist}(X,F)$.