<s>[INST] Find the locus of the vertices of the right circular cones that pass through the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$ 
Prove that the locus of the vertices of the right circular cones that pass through the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$ is $\frac{x^2}{a^2-b^2}-\frac{z^2}{b^2}=1, y=0$ or $\frac{y^2}{a^2-b^2}+\frac{z^2}{a^2}=-1, z=0$.  

EDIT:
Here $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$ is the base of the cone. Let the vertex of the cone be $(x_1,y_1,z_1)$. Let the generator be $$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}.$$
Then $z=0$ implies any point on ellipse be $(x_1-lz_1/n, y_1-mz_1/n,0)$. It lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then the point satisfies the ellipse we get $$\frac{(nx_1-lz_1)^2}{a^2}+\frac{(ny_1-mz_1)^2}{a^2}=n^2$$ 
Eliminating $l,m,n$ we get the equation of the cone as $$\frac{1}{a^2}(zx_1-xz_1)^2+\frac{1}{b^2}(zy_1-yz_1)^2=(z-z_1)^2.$$ How to get the locus of vertex $(x_1,y_1,z_1)$?
Edit 2
How to get the locus of the vertices in the given two forms using purely mathematical way? 
 I not able to solve the problem. Please help. 
 [/INST] Considering that


*

*the ellipse on the $x,y$ plane is symmetrical with repect to both axis, and also it is symmetrical with respect to the $z,x$ and $z,y$ planes.

*a circular cone whose axis is incident to the $x,y$ plane, will intercept on it a quadric that is symmetrical to the plane containing the cone axis and parallel to the $z$ axis. 


we conclude that the axis of the cone shall lie either on the $z,x$ or in the $z,y$ plane, and so shall thus be for the vertex .  
case: cone axis in the $z,x$ plane 
Let the vertex be $V=(x_v,0,z_v)$, and the unit vector defining the axis $\mathbb {a} =(cos \alpha, 0, sin\alpha)$.
A point $P$ on the circular cone will obey to
$$
\frac{{\mathop {VP}\limits^ \to  }}
{{\left| {\mathop {VP}\limits^ \to  } \right|}} \cdot \mathbf{a} = const. = \cos \beta 
$$
and in particular a point on the $x,y$ plane will obey to:
$$
\left( {\left( {x - x_{\,v} } \right)\cos \alpha  + \left( {0 - z_{\,v} } \right)\sin \alpha } \right)^{\,2}  = \left( {\left( {x - x_{\,v} } \right)^{\,2}  + \left( {0 - z_{\,v} } \right)^{\,2}  + y^{\,2} } \right)\cos ^{\,2} \beta 
$$
that is (duly excluding the cases $\beta=\pi /2$ and $\beta=\alpha$ which can be dealt apart)
$$
\begin{gathered}
  \left( {z_{\,v} ^{\,2}  + y^{\,2} } \right)\cos ^{\,2} \beta  =  \hfill \\
   = \left( {\cos ^{\,2} \alpha  - \cos ^{\,2} \beta } \right)\left( {x - x_{\,v}  + z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \alpha  - \cos \beta } \right)}}} \right)\left( {x - x_{\,v}  + z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \alpha  + \cos \beta } \right)}}} \right) \hfill \\ 
\end{gathered} 
$$
$$
\left( {1 - \frac{{\cos ^{\,2} \alpha }}
{{\cos ^{\,2} \beta }}} \right)\left( {x - x_{\,v}  - z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \beta  - \cos \alpha } \right)}}} \right)\left( {x - x_{\,v}  + z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \beta  + \cos \alpha } \right)}}} \right) + y^{\,2}  =  - z_{\,v} ^{\,2}  \tag{1}
$$
First step for this to represent the targeted ellipse is to get rid of the term in $x^1$, which means
that is shall be:
$$
 - x_{\,v}  - z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \beta  - \cos \alpha } \right)}} =  - \left( { - x_{\,v}  + z_{\,v} \frac{{\sin \alpha }}
{{\left( {\cos \beta  + \cos \alpha } \right)}}} \right)
$$
i.e.
$$
 - z_{\,v} \left( {\frac{{\sin \alpha \;\cos \alpha }}
{{\cos ^{\,2} \beta  - \cos ^{\,2} \alpha }}} \right) = x_{\,v}  \tag{2}
$$
So that eq. (1) can be rewritten as:
$$
\left( {1 - \frac{{\cos ^{\,2} \alpha }}
{{\cos ^{\,2} \beta }}} \right)\left( {x + x_{\,v} \frac{{\cos \beta }}
{{\cos \alpha }}} \right)\left( {x - x_{\,v} \frac{{\cos \beta }}
{{\cos \alpha }}} \right) + y^{\,2}  =  - z_{\,v} ^{\,2} 
$$
leading to:
$$
\begin{gathered}
  \left( {1 - \frac{{\cos ^{\,2} \alpha }}
{{\cos ^{\,2} \beta }}} \right)x^{\,2}  + y^{\,2}  = x_{\,v} ^{\,2} \frac{{\left( {\cos ^{\,2} \beta  - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }} - z_{\,v} ^{\,2}  =  \hfill \\
   = x_{\,v} ^{\,2} \left( {\frac{{\left( {\cos ^{\,2} \beta  - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }} - \frac{{\left( {\cos ^{\,2} \beta  - \cos ^{\,2} \alpha } \right)^{\,2} }}
{{\sin ^{\,2} \alpha \;\cos ^{\,2} \alpha }}} \right) =  \hfill \\
   = x_{\,v} ^{\,2} \frac{{\left( {\cos ^{\,2} \beta  - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }}\frac{{1 - \cos ^{\,2} \beta }}
{{1 - \cos ^{\,2} \alpha }} \hfill \\ 
\end{gathered}  \tag{3}
$$
Comparing this to the target we obtain
$$
\left\{ \begin{gathered}
  x_{\,v} ^{\,2} \frac{{\left( {\cos ^{\,2} \beta  - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }}\frac{{\sin ^{\,2} \beta }}
{{\sin ^{\,2} \alpha }} = b^{\,2}  \hfill \\
  \left( {\frac{{\cos ^{\,2} \beta  - \cos ^{\,2} \alpha }}
{{\cos ^{\,2} \beta }}} \right) = \frac{{b^{\,2} }}
{{a^{\,2} }} \hfill \\ 
\end{gathered}  \right.
$$
With some algebraic manipulation
$$
\left\{ \begin{gathered}
  x_{\,v} ^{\,2} \frac{{\left( {\cos ^{\,2} \beta  - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }}\frac{{\sin ^{\,2} \beta }}
{{\sin ^{\,2} \alpha }} = b^{\,2}  \hfill \\
  x_{\,v} ^{\,2} \frac{{\cos ^{\,2} \beta }}
{{\cos ^{\,2} \alpha }}\frac{{\sin ^{\,2} \beta }}
{{\sin ^{\,2} \alpha }} = a^{\,2}  \hfill \\
  x_{\,v} ^{\,2} \frac{{\sin ^{\,2} \beta }}
{{\sin ^{\,2} \alpha }} = a^{\,2}  - b^{\,2}  \hfill \\
  \frac{{\left( {\cos ^{\,2} \beta  - \cos ^{\,2} \alpha } \right)}}
{{\cos ^{\,2} \alpha }} = \frac{{b^{\,2} }}
{{\left( {a^{\,2}  - b^{\,2} } \right)}} \hfill \\
  \left( {\frac{{\cos ^{\,2} \beta  - \cos ^{\,2} \alpha }}
{{\cos ^{\,2} \beta }}} \right) = \frac{{b^{\,2} }}
{{a^{\,2} }} \hfill \\
  \frac{{\cos ^{\,2} \beta }}
{{\cos ^{\,2} \alpha }} = \frac{{a^{\,2} }}
{{\left( {a^{\,2}  - b^{\,2} } \right)}} \hfill \\ 
\end{gathered}  \right.
$$
we finally arrive to rewrite the eq. (2) as:
$$
z_{\,v} ^{\,2}  = \frac{{b^{\,2} }}
{{\left( {a^{\,2}  - b^{\,2} } \right)}}\left( {1 - \frac{{\sin ^{\,2} \beta }}
{{\sin ^{\,2} \alpha }}} \right)x_{\,v} ^{\,2}  = \frac{{b^{\,2} }}
{{\left( {a^{\,2}  - b^{\,2} } \right)}}x_{\,v} ^{\,2}  - b^{\,2} 
$$
$$
\frac{{x_{\,v} ^{\,2} }}
{{\left( {a^{\,2}  - b^{\,2} } \right)}} - \frac{{z_{\,v} ^{\,2} }}
{{b^{\,2} }} = 1
$$
case: cone axis in the $z,y$ plane
The overall scheme is the same, just we have to exchange $x$ with $y$ and $a$ with $b$, giving
$$
\frac{{y_{\,v} ^{\,2} }}
{{\left( {b^{\,2}  - a^{\,2} } \right)}} - \frac{{z_{\,v} ^{\,2} }}
{{a^{\,2} }} = 1
$$
or if you prefer:
$$
\frac{{y_{\,v} ^{\,2} }}
{{\left( {a^{\,2}  - b^{\,2} } \right)}} + \frac{{z_{\,v} ^{\,2} }}
{{a^{\,2} }} =  - 1
$$
</s>