Five points on $(x-y)$ plane $ (x1,y1), (x2,y2),… (x5,y5)$ defining a common conic intersection curve $C$ are obtained by section with variably inclined right circular cones i.e.,variable vertex $V$ position.
Find locus $L$ of all vertices $V.$
Background:
We have $ \epsilon$= eccentricity , $\gamma=$ semi-vertical angle of cone, $\alpha =$ angle to symmetry axis and since
$$\epsilon=\dfrac{\cos\gamma}{\cos\alpha},$$
the above relation implies that that every conic section is associated uniquely with an intersected right circular cone upto
- scale
- rotation around axis of symmetry and
- translation
Special case : If $C$ is a circle then $L$ is the common axis of symmetry of each cone.
Thanks in advance for pointers to vertex locus finding.
Best Answer
If the given conic is either an ellipse or a hyperbola, vertex $V$ of the cone must lie on the plane perpendicular to the plane of the conic and passing through its foci.
Let $A$ and $B$ be the endpoints of the transverse axis of the conic, $V$ the cone vertex, and $2u$ the aperture angle of the cone. Let then $P$ be any point on the ellipse and $H$ its projection onto $AB$. A plane through $P$, perpendicular to the axis of the cone, intersects the cone along a circle $A'B'P$ (see diagram below), where $A'$ and $B'$ lie on plane $VAB$.
By the intersecting chords theorem we know that $PH^2=A'H\cdot B'H$. But, on the other hand, we have by similitude: $$ A'H:AH=BC:AB, \quad\hbox{that is:}\quad A'H={BC\over AB}\cdot AH; $$ $$ B'H:BH=AD:AB, \quad\hbox{that is:}\quad B'H={AD\over AB}\cdot BH. $$
If we set $n=VA$, $m=VB$ and $2a=AB$, the above formulas can be written as $$ A'H={m\sin u\over a}\cdot AH, \quad B'H={n\sin u\over a}\cdot BH, $$ and inserting these into the formula for $PH^2$ we get: $$ PH^2={mn\sin^2 u\over a^2}\,AH\cdot BH. $$ If $H$ is a focus of the conic, then $PH={b^2/a}$ is the semi-latus rectum, while $AH\cdot BH=|a^2-c^2|=b^2$ (as usual $2b$ is the length of the conjugate axis and $2c$ is the distance between foci) and from the above equation we get: $$ \tag{1} mn={b^2\over \sin^2 u}. $$ Another equation for $m$ and $n$ can be found from the cosine rule applied to triangle $AVB$: $$ \tag{2} m^2+n^2\mp2mn\cos2u=4a^2, $$ where sign $-$ must be taken for an ellipse ($\angle AVB=2u$) and sign $+$ for a hyperbola ($\angle AVB=\pi-2u$).
If the conic is an ellipse one obtains from $(1)$ and $(2)$ $$(m-n)^2=4(a^2-b^2),$$ that is the difference of the distances of $V$ from $A$ and $B$ is constant. The locus of $V$ is then a hyperbola, having its foci at the vertices of the ellipse and semi-major axis $\sqrt{a^2-b^2}$. This hyperbola thus passes through the foci of the ellipse.
If the conic is a hyperbola you get instead $$(m+n)^2=4(a^2+b^2)$$ and the locus of $V$ is an ellipse having its foci at the vertices of the hyperbola and semi-major axis $\sqrt{a^2+b^2}$. This ellipse thus passes through the foci of the hyperbola.
Finally, if the conic is a parabola of vertex $A$ and semi-latus rectum $p$ one gets from a similar reasoning: $$ AV=n={p\over2\sin^2u}. $$ From that it follows that the distance of vertex $V$ from the plane of the parabola is $$VK=n\sin2u=p\cot u,$$ while the distance from the projection of $V$ on that plane to the focus of the parabola is $$KF=n\cos2u+p/2={p\over2}\cot^2u.$$ Hence $KF=VK^2/(2p)$ and the vertex thus lies on a a parabola, equal to the given parabola and having its vertex at the focus of the latter.
We can summarise the above results as follows:
You can see below an example: the green ellipse on plane x-y has vertices $A$, $B$, foci $F$, $G$ and eccentricity $1/\sqrt2$. The locus of the vertex $V_1$ of any cone intercepting the ellipse is the pink hyperbola on plane x-z, with vertices $F$, $G$, foci $A$, $B$ and eccentricity $\sqrt2$. Conversely, the ellipse is the locus of the vertex $V_2$ of any cone intercepting the hyperbola. Note also that the axes of the cones are tangent to the locus where the vertex lies.