Solution
= Solution
Let
$$
\widehat\beta=(X^TX)^{-1}X^TY,
\qquad
\widehat\sigma^2
=\frac{\|Y-X\widehat\beta\|^2}{n-p}.
$$
The pivotal quadratic form has an $F_{p,n-p}$ distribution, so the $(1-\alpha)$ <normal linear-model confidence ellipsoid> is
$$
\boxed{
\left\{b\in\mathbb R^p:
\frac{(b-\widehat\beta)^TX^TX(b-\widehat\beta)}
{p\widehat\sigma^2}
\leq F_{p,n-p}(1-\alpha)
\right\}
}.
$$
Here $F_{p,n-p}(1-\alpha)$ is the $(1-\alpha)$ quantile of the indicated $F$ distribution.