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. 

You have a good start of getting an enveloping elliptic cone:
$$\frac{(Z x-X z)^2}{a^2}+\frac{(Y z-Z y)^2}{b^2}=(z-Z)^2$$
Rearrange to

$$
\begin{pmatrix}
  x & y & z
\end{pmatrix}
\begin{pmatrix}
  \frac{Z^2}{a^2} & 0 & -\frac{XZ}{a^2} \\
  0 & \frac{Z^2}{b^2} & -\frac{YZ}{b^2} \\
  -\frac{ZX}{a^2} & -\frac{ZY}{b^2} & \frac{X^2}{a^2}+\frac{Y^2}{b^2}-1
\end{pmatrix}
\begin{pmatrix}
  x \\ y \\ z
\end{pmatrix}+
2Zz-Z^2=0$$

Due to circular symmetry of right circular cone, the quadratic terms can be diagonalized into the form of $$\lambda x'^2+\lambda y'^2+\mu z'^2$$
which means the matrix has two equal eigenvalues.
That is
$$\det
\begin{pmatrix}
  \frac{Z^2}{a^2}-\lambda & 0 & -\frac{XZ}{a^2} \\
  0 & \frac{Z^2}{b^2}-\lambda & -\frac{YZ}{b^2} \\
  -\frac{ZX}{a^2} & -\frac{ZY}{b^2} & \frac{X^2}{a^2}+\frac{Y^2}{b^2}-1-\lambda
\end{pmatrix}=0$$
has a double root.

Also the vertex should lie on the mirror plane of the ellipse (i.e. either $X=0$, $Y=0$ or $Z=0$; that's only the case for right circular cone).

Case I: $X=0$
$$
\left( \frac{Z^2}{a^2}-\lambda \right)
\left[
  \lambda^2+
  \left( 1-\frac{Y^2+Z^2}{b^2} \right)\lambda-
  \frac{Z^2}{b^2}
\right]=0$$
We have two positive roots and one negative root, both factors share the same root $\lambda=\dfrac{Z^2}{a^2}$.
Now
$$\frac{Z^4}{a^4}+
  \left( 1-\frac{Y^2+Z^2}{b^2} \right) \frac{Z^2}{a^2}-
  \frac{Z^2}{b^2}=0 $$

$$\frac{Y^2}{a^2-b^2}+\frac{Z^2}{a^2} = -1$$
which is an imaginary ellipse provided $a^2 > b^2$.

Case II: $Y=0$
$$
\left( \frac{Z^2}{b^2}-\lambda \right)
\left[
  \lambda^2+
  \left( 1-\frac{X^2+Z^2}{a^2} \right)\lambda-
  \frac{Z^2}{a^2}
\right]=0$$
We have two positive roots and one negative root, both factors share the same root $\lambda=\dfrac{Z^2}{b^2}$.
Now
$$\frac{Z^4}{b^4}+
  \left( 1-\frac{X^2+Z^2}{a^2} \right) \frac{Z^2}{b^2}-
  \frac{Z^2}{a^2}
  =0 $$

$$\frac{X^2}{a^2-b^2}-\frac{Z^2}{b^2} = 1$$
  which is a hyperbola provided $a^2 > b^2$.

Case III: $Z=0$
$$\lambda^2
  \left( \frac{X^2}{a^2}+\frac{Y^2}{b^2}-1-\lambda \right) = 0$$
The double root is $\lambda=0$, the cone degenerates to a flat plane $z=0$.

The vertex is any point on
$$Z=0$$

Further points to be noticed:
The conics $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \cap z=0$ and $\frac{x^2}{a^2-b^2}-\frac{z^2}{b^2}=1  \cap y=0$ are known as the focal ellipse and focal hyperbola of confocal quadrics
$$\frac{x^2}{a^2+t}+\frac{y^2}{b^2+t}+\frac{z^2}{t}=1$$
where $a^2>b^2$.
See also another answer with picture of mine
.
