Question on when a section of a zero set of a polynomial becomes a submanifold? $\mathbf {The \ Problem \ is}:$ If $p : \mathbb R^{m+n} \to \mathbb R$ with $p(z)= \sum_{i=1}^m {a_i}^2{x_i}^2 - \sum_{i=1}^n {b_j}^2{y_j}^2$ ,$($$z=(x_1,...,x_m,y_1,..,y_n)$$)$
A) Then for what values of $a_i , b_j$ the set $Z = p^{-1}(0) \cap S^{m+n-1}$ a manifold ?
B) Also if $a_i =1=b_j$ for all $i,j$ show $Z$ is a connected manifold iff $m,n \geq 2$, and find it's homotopy type .
$\mathbf {My \  approach}:$ Here, $p(z)$ is homogenous, so $0$ is the only critical value of $p.$
Now, for B), only if part is ok, but I'm unable to do if part .
If we can construct another smooth map $f$ on $S^{m+n-1}$ such that $f^{-1}(0)=Z$ with $0$ being a regular value of $f$, then it's done .
A small hint is warmly appreciated for both A) and B).
 A: For any a diagonal matrix $A$ define
$$ p_A(z) = z^\top A z = \left< z, z\right>_A$$
Where $\left< . , . \right>_A$ is the bilinear form defined by $A$.
You have $D_z p (v) = 2 \left< v, z\right>_A$ for $v \in T_z \mathbb{R}^{n+m} = \mathbb{R}^{n+m}$. We have $S^{n+m-1} = p_E^{-1}(0)$ and $T_z S^{n+m-1} = \ker D_zp_I$ with $E$ the identity matrix.
Now take $ A = diag(a_1^2,...,a_m^2,-b_1^2,...,-b_n^2)$ and define
$$f : \mathbb{R}^{m+n} \to \mathbb{R}^2\\
z \mapsto (p_A(z), p_I(z)-1)$$
You have $Z= f^{-1}(0) = S^{n+m-1} \cap p_A^{-1}(0)$. Let $z\in Z$, then $D_z f = (D_z p_A, D_z p_I)^\top = 2(Az,z)^\top$ is surjective if and only if the vectors $z$ and $Az$ are linearly independent, which holds if and only if $z$ is not a eigenvector of $A$. If $z\in Z$ fulfills $Az = \lambda z$ for some $\lambda \in \mathbb{R}$, we have $0 = p_A(z) = \lambda p_I(z) = \lambda $. So if all $a_i, b_j$ are nonzero we are fine. If one of the values is zero, I believe you can consider $\epsilon$-balls around $z \in S^{m+n-1}\cap \ker A$  and proof that their intersection with $Z$ is not contractible, but I didn't do the details. I also don't know how to tackle the homotopy type of $Z$.
