I was trying to solve some change of basis exercises, so I've done them and I want you to tell me if I've done it correctly.
1) Given the coordinates system $(O'; X'' Y'')$ asociated to the basis $B=[{b_1=\frac{1}{\sqrt{2}}; \frac{1}{\sqrt{2}}; > b_2=\frac{-1}{\sqrt{2}}; \frac{1}{\sqrt{2}}}]$ with origin in $O'=(2;5)$, find the intersection between $Q: y''^2=-4 \sqrt{2} x''$ and the straight line in $(O,XY)$ $V=(0;5)+t(-1;1) t\epsilon R$
Well, first I tried to write $Q$ in terms of $X$ and $Y$, and not in terms of $X''$ and $Y''$. I used the formula: $(\begin{matrix} X''\\ Y'' \\ \end{matrix})$ =$(C)_B$ × $(\begin{matrix} X-2\\ Y-5 \\ \end{matrix})$
Then I replaced those values in the equation of $Q$ and I got $Q: \frac{(-x+y-3)^2}{2}=-4 \sqrt{2} \frac{(x+y-7)}{\sqrt{2}}$
After that, I found the intersection between $Q$ and $V$ using equations and I got that the intersection points in $(O;XY)$ are $P_1(3;2)$ and $P_2(-1;6)$
2)Given the coordinates system $(O,X'Y')$ asociated to the basis $B=[{b_1=\frac{1}{\sqrt{2}}; \frac{1}{\sqrt{2}}; > b_2=\frac{-1}{\sqrt{2}}; \frac{1}{\sqrt{2}}}]$ with origin in $O=(0;0)$ and the ellipse $9X'^2+5Y'^2-3X'+2Y'-10=0$ find the coordinates of the centre of the ellipse in $(O,XY)$
So, here I found the centre of the ellipse in $(O,X'Y')$ and it was $(\frac{1}{6}; - \frac{1}{5})$, then, doing something similar to what I've done before I found out that the centre in $(O,XY)$ was $( \frac{11}{30 \sqrt{2}};$ $\frac{-1}{30 \sqrt{2}}$)