<s>[INST] The quadratic equation with two unknowns: $ x^2 - 2axy + by^2 = 0 $ Please, could you help me with the next problem:
The problem:
Determine real parameters $a, b \in \Bbb{R}$, such that with: 
$$ \langle x, y \rangle = x_1y_1 - ax_2y_1 - ax_1y_2 + bx_2y_2 $$ 
the inner product is defined in (or is it 'on') the vector space $\Bbb{R}^2$.
1) In order to $\langle x, x \rangle \ge 0$, i got that it should be $a \in \left[ -1, 1\right]$ and $b \ge 1$.
2) In order to $\langle x, x \rangle = 0 \,\Leftrightarrow\, x = 0$, i came to the equation $x^2 - 2axy + by^2 = 0$ (x = x_1, y= x_2).
My solution is next:
$$x^2 - 2axy + by^2 = 0 \,\Leftrightarrow\, x^2 + by^2 = 2axy \,\Leftrightarrow\, \frac{x^2}{y^2} + b = \frac{2ax}{y} \,\Leftrightarrow\, $$
$$\,\Leftrightarrow\, b = \frac{2ax}{y} - \frac{x^2}{y^2} \,\Leftrightarrow\, \frac{2ax}{y} - \frac{x^2}{y^2} \ge 1 \,\Leftrightarrow\, \frac{2ax}{y} \ge \frac{x^2}{y^2} \setminus :\frac{y^2}{x^2} \,\Leftrightarrow\,$$
$$\,\Leftrightarrow\, \frac{2ay}{x} \ge 1.$$
Now, because $x$ and $y$ are arbitrary, it's easy to find such $x$ and $y$, that for those numbers final inequality isn't correct. 
So, my conclusion is that there isn't solution for this equation. Am i right?
Thank you, in advance, for your time!
P.S. I'm sorry for not writing the whole problem in the first place. I just wasted my and yours time.
 [/INST] Here's an answer building on the comments.
\begin{align}
\langle x,x \rangle &= x_1^2 - 2ax_1x_2 + bx_2^2 \\
&= x_1^2 - 2ax_1x_2 + a^2x_2^2 + (b-a^2)x_2^2 \\
&= \underbrace{(x_1 - ax_2)^2}_{\geq 0} + \underbrace{(b-a^2)x_2^2}_{\geq 0 \text{ if }a^2\leq b}
\end{align}
So we are guaranteed that $\langle x,x \rangle\geq 0$  for all $x\in\mathbb{R}^2$ if $a^2 \leq b$.
Next, we can write:
\begin{align}
\langle x,x \rangle &= 0\\
x^2_1 - 2ax_1x_2 + bx_2^2 &= 0 \\
\frac{x_1^2}{x_2^2} - 2a\frac{x_1}{x_2} + b &= 0 \\
z^2 - 2az + b &= 0 \\
z = \frac{1}{2}\left( 2a\pm \sqrt{4a^2 - 4b} \right) &= \frac{x_1}{x_2}\\
x_2\left(a\pm\sqrt{a^2-b}\right) &= x_1
\end{align}
since $x=(x_1,x_2)\in\mathbb{R}^2$, we have $a^2\geq b$.
But since $a^2 \geq b$ and $a^2 \leq b$, we get $a^2 = b$.
</s>