Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ with $n \geq 2$ defined by $$f(x) = \langle x,x \rangle ^2 - \langle a,x \rangle ^2,$$ where $a \in \mathbb{R}^n \setminus \{0\}$. Prove that $f$ has two local minima and calculate them.
I know that I have to use the maximum-minimum theorem on $\overline{B}(0; \|a \|) := \{x \in \mathbb{R}^n \mid \|x \| \leq \|a\| \}$, but I don't know how I can use this to prove that $f$ has two local minima.
Hint.
As in the stationary points we have
$$ \nabla f = 4(x\cdot x)x - 2(a\cdot x)a = 0 $$
making $x=\lambda a$ and substituting
$$ \nabla f = 4\lambda^3(a\cdot a)a - 2\lambda (a\cdot a)a = \lambda(4\lambda^2-2)(a\cdot a)a = 0 $$
which gives
$$\lambda = \cases{0\\ \pm\frac{\sqrt{2}}{2}} $$
so the points to test are
$$ x^{*}= \pm\frac{\sqrt{2}}{{2}}a $$
NOTE
For $\lambda = 0$ we have $\det\left(\nabla^2 f\right) = 0 $.
Considering now $x_{\delta} = x^* +\delta = \frac{a}{\sqrt{2}}+\delta$ we have that near $x^*$
$$ f(x_{\delta}) \ge f(x^*) $$
because
$$ f(x_{\delta}) = \left((\frac{a}{\sqrt{2}}+\delta)\cdot(\frac{a}{\sqrt{2}}+\delta)\right)^2-\left((\frac{a}{\sqrt{2}}+\delta)\cdot a\right)^2 = -\frac{(a\cdot a)^2}{4}+(a\cdot\delta)^2+(a\cdot a)(\delta\cdot\delta)+2\sqrt{2}(a\cdot\delta)(\delta\cdot\delta)+(\delta\cdot\delta)^2 $$
Note that $\exists\delta > 0$ such that $ (a\cdot a) \ge 2\sqrt{2} (a\cdot\delta)$