So , i have been learning about conics .
My textbook gives three theorms
1) Cutting a double cone by a plane in any way , you would get curve , such that distance between any point of the curve from a fixed point is propotional to distance between the point and a fixed line
2)eccentricity of the curve is = $\frac{cos(\alpha)}{cos(\beta)}$ , where $\alpha$ is the angle between cutting plane and axis , $\beta$ is the angle between axis and cone.
3) Any 2 deegre curve with 2 variable represent one of the curves obtained above
I know these theorms are pretty basic , but i have not been able to prove them . I tried to find it on internet , but unsuccessful. Pls help me , links , partial solution anything are appreciated .
Best Answer
Let $B$ be the apex of a right circular cone. Let $\beta$ be the angle made between a generatrix of the cone and its axis. Let $\mathscr{P}$ be a plane not containing $B$ and intersecting the cone at an angle $\alpha$ with the axis. Let $\mathscr{Q}$ be the plane containing the axis and perpendicular to $\mathscr{P}$.
Construct the Dandelin sphere tangent to $\mathscr{P}$ and each generatrix of the cone. If $\alpha>\beta$, its center will be the incenter of the triangle cut in $\mathscr{Q}$ by $\mathscr{P}$ and the cone; if $\alpha<\beta$, it will be the excenter. (The case $\alpha=\beta$ can be handled as the excenter.)
Let $F$ be the point of tangency $\mathscr{P}$ with the sphere. Let $\mathscr{R}$ be the plane containing the circle of intersection between the sphere and the cone. Let $m$ be its line of intersection with $\mathscr{P}$. I claim that the section cut in the cone by $\mathscr{P}$ is the conic with focus $F$, directrix $m$, and eccentricity $\varepsilon=\frac{\cos\alpha}{\cos\beta}$.
To see this, let $A$ be a point on the section. Let $C$ be the foot of the perpendicular from $A$ to $m$. Let $D$ be the foot of the perpendicular from $A$ to $\mathscr{R}$, and let $E$ be the point where the generatrix through $A$ intersects the circle of intersection.
Let $x = AD$. Note that the measure of $\angle DAE$ is $\beta$ and that the measure of $\angle DAC$ is $\alpha$. Because $\overline{AF}$ and $\overline{AE}$ are both tangent to the sphere, $AE=AF$. So $\cos\beta=\frac{x}{AF}$. On the other hand, $\cos\alpha=\frac{x}{AC}$. So $\frac{\cos\alpha}{\cos\beta}=\frac{AF}{AC}$.
It follows that ratio of $AF$ to the distance from $A$ to $m$ is constant, and equal to $\varepsilon=\frac{\cos\alpha}{\cos\beta}$. So $A$ is on the conic with focus $F$, directrix $m$, and eccentricity $\varepsilon=\frac{\cos\alpha}{\cos\beta}$.