Find values of the parameter for the given conic Given $\mathbb{R}^2$ an affine space and the conics:
$Q_\alpha:3x_1^2-\alpha x_1x_2+3x_2^2+14x_1-2x_2+3=0$
$C_\beta:x_1^2+2x_2^2+2\beta x_1x_2-6x_1+5=0$
$i)$ Find $\alpha$ and $\beta$ such that $Q_\alpha$ is a hyperbola and  the line
$(r):x_2=x_1-1$ is tangent to $C_\beta$
$ii)$ For the value found for $\beta$ at $i)$ find the canonical form of $C_\beta$ (i.e., classify the conic) using isometries
My Attempt:
I would like to know whether my approach is right and if otherwise, how to correct it.
$i)$
For the first parameter: $\alpha$
In order to find $\alpha$ i thought it would be useful to require that $\Delta≠0$ and $\delta<0$
Where if $Q_\alpha:x^\top A x+Bx+c=0, \delta=det(A)$ and $\Delta=det(A_1)$ where
$\begin{align*}A_1= \begin{bmatrix}
                A & \frac{1}{2}B^\top \\
                \frac{1}{2}B & c\\
                \end{bmatrix}
\end{align*}, $ but it does not help me too much as I have already seen .
For the other parameter: $\beta$
I would just plug in the value of $x_2=x_1-1$ in the equation of $C_\beta$ and require the discriminant to be zero but this method does not always  work  as I have been told(I think that in dimension 3 things do not work this way? Given that we are working with the asymptote cone). How should I proceed in this situation?
$ii)$
In order to classify the conic using isometries I would try to find the rotation R(an orthogonal matrix with $det(R)=1$) of $\mathbb{R}^2$ making the change of coordinates $x=Rx'$ for x $\in\mathbb{R}^2$ in the equation of $C_\beta:x^\top A' x+B'x+c'=0$ using eigenvectors and eventually Completing the Squares if needed.
Another Question: How could this kind of problems be tackled in a simple way? I mean finding the parameter from the equation of a quadric surface knowing its nature(i.e. an ellipse, or a hyperboloid of one sheet for quadric surfaces) I have seen similar problems here but their solution seems too complicated, having to know some previous classifications and so on.

i) Matrices associated with the first, resp. second, conic curve are:
$$A=\begin{pmatrix}3&-\frac12 \alpha & 7\\
-\frac12 \alpha & 3&-1\\7&-1&3\end{pmatrix} \ \& \ B=\begin{pmatrix}1&\beta&-3\\ \beta&2&0\\-3&0&5 \end{pmatrix} $$
The first conic, as you have said, will be a hyperbola when $\delta=9-\frac14 \alpha^2 <0$, i.e., when
$\alpha >6 $ or $\alpha < -6$.
The constraint of tangency for the second conic is dealed with by  working by duality : compute the inverse of $B$, or more exactly the adjunct matrix of $B$, which is the matrix of the so called dual conic:
$$B'=\begin{pmatrix}10&-5\beta&6\\-5\beta&-4&-3\beta\\6&-3\beta&(2-\beta^2)\end{pmatrix} $$
and express that the vector $V^T=(-1 \ \ 1 \ \ 1)^T$ of the coefficients of the straight line belongs to this dual conic by writing equation:
$$V^TB'V=0 \iff (-1 \ \ 1 \ \ 1)\begin{pmatrix}10&-5\beta&6\\-5\beta&-4&-3\beta\\6&-3\beta&(2-\beta^2)\end{pmatrix}\begin{pmatrix}-1\\1\\1\end{pmatrix}=0$$
giving a quadratic equation
$$ -\beta^2+4\beta-4=0$$
whose unique double root is
$$\beta=2.$$

Fig. : A graphical representation of the conic curves for the cases $a=7$ and $b=2$, the tangency point (black dot) being $(1,0)$.
