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.
 A: 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 $(1,0)$.
