Ich würde mir das erstmal anschauen z.B. in GeoGebra....
zweckslesbakeit
-x²-y²+αz²-6xy-4x+4y-2αz=α
===> \(\small A \, := \, \left(\begin{array}{rrr}-1&-3&0\\-3&-1&0\\0&0&\alpha\\\end{array}\right) , a^T \, := \, \left\{ -4, 4, 2 \; \alpha \right\} , a_0 \, := \, \alpha\)
JD:=JordanDiagonalization(A) ===>
\(\small JD \, := \, \left\{ \left(\begin{array}{rrr}-1&1&0\\1&1&0\\0&0&1\\\end{array}\right), \left(\begin{array}{rrr}2&0&0\\0&-4&0\\0&0&\alpha\\\end{array}\right) \right\} \)
===> Drehung/Rotation R
\(\small R \, := \, \left(\begin{array}{rrr}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\0&0&1\\\end{array}\right)\)
q_D(X):=(x,y,z) D (x,y,z) + a R (x,y,z) = α
\(q_D: \, \alpha \; z^{2} + 2 \; x^{2} - 4 \; y^{2} + 2 \; \alpha \; z - 4 \; \sqrt{2} \; x = \alpha\)
Translation
\(T \, := \, \left\{ x = x + \sqrt{2}, y = y, z = z - 1\right\} \)
q_D(T)
\(q_N: \, \alpha \; z^{2} + 2 \; x^{2} - 4 \; y^{2} = 2 \; \alpha + 4\)
Kommst Du nun zurecht?