# Which points in the given plane can be tangent to some (hyper-)ellipsoid?

Write $\mathbb{R}^N=\{(x_1,x_2,...,x_N)|\ x_j \in \mathbb{R} \ \forall \ j=1,...,N\}$ and consider a point $a$ in the hyperplane $x_1=1$. Does there exist a hyper-ellipsoid $E_a$, with non-zero volume in $\mathbb{R}^N$ and centered at the origin such that the hyperplane $x_1=1$ is tangent to $E_a$ at $a$?

Work so far:

This is true for $N=2$, since, in $\mathbb{R}^2$, $x^2+bxy+cy^2=1$ is the equation for an ellipse provided $b^2 \leq 4ac$. For the point $a=(1,y_0)$, with $y_0 \neq 0$, the ellipse

$2x^2 - \dfrac{2xy}{y_0} + \dfrac{y^2}{y_0^2}=1$

is centered at the origin and is tangent to the line $x=1$ at $y_0$. If $y_0=0$, the unit circle works. Does this property generalise for $N>2$?

Extending your example, the ellipsoid $$2x_1^2 - \frac{2x_1x_2}{t} + \frac{x_2^2}{t^2} + x_3^2 + x_4^2 + \dots + x_n^2 = 1$$ is centered at the origin and tangent to the hyperplane $x_1 = 1$ at $P = (1,t,0,0,\dots,0,0)$. We may then apply a rotation to the last $n-1$ coordinates which will preserve the hyperplane but move the point $P$ to an arbitrary point that's a distance $t$ away from $(1,0,0,\dots,0,0)$.
• Thanks! I suppose this can be extended to find an ellipsoid tangent to any (nonzero) point a distance $k$ from the origin in $\mathbb{R}^N$ by finding $t$ such that $\sqrt{(1+t^2)}=k$, applying a rotation to the first two coordinates until the point $(1,t,0,...,0)$ is mapped to the correct $x_1$ coordinate, and then applying a similar transformation to the one you suggest to the last $n-1$ coordinates to hit the desired point(?) Commented Mar 21, 2017 at 17:09
• Sure. Or another way: if you wanted to be tangent to the plane $x_1 = p_1$ at $(p_1,p_2,\dots,p_n)$, you could take the ellipse tangent to $x_1 = 1$ at $(1, \frac{p_2}{p_1}, \dots, \frac{p_n}{p_1})$, and then scale it by a factor of $p_1$. Commented Mar 21, 2017 at 17:28