# (10%) In R^2, there are two sets of coordinate systems, represented by two distinct bases: (x_1, y_1) and (x_2, y_2). If the equations of the same ellipse represented

(10%) In ${R}^{2}$, there are two sets of coordinate systems, represented by two distinct bases: $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$. If the equations of the same ellipse represented by the two distinct bases are described as follows, respectively: $2\left({x}_{1}{\right)}^{2}-4\left({x}_{1}\right)\left({y}_{1}\right)+5\left({y}_{1}{\right)}^{2}-36=0$ and $\left({x}_{2}{\right)}^{2}+6\left({y}_{2}{\right)}^{2}-36=0.$ Find the transformation matrix between these two coordinate systems: $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$.

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Let the transformation matrix be $A=\left[\begin{array}{cc}P& Q\\ R& S\end{array}\right]$

Putting in thhe given EQN. Since the conversion is from a non homogeneus to homogeneous QN.,

Which is obtained by rotation through angle tyou can use the following transformation matrix as a formula $\left[\begin{array}{cc}COS\left(T\right)& SIN\left(T\right)\\ -SIN\left(T\right)& COS\left(T\right)\end{array}\right]\left[\begin{array}{c}X1\\ Y1\end{array}\right]=\left[\begin{array}{c}X2\\ Y2\end{array}\right]$

Where $T=\frac{1}{2}{\mathrm{tan}}^{-1}\left(\frac{2H}{A-B}\right)$

Where

$Y{1}^{2}=5$

Solving we get $P=0.89443\dots \dots Q=0.44721\dots \dots R=--0.44721\dots \dots \dots S=0.89443$

Hence the transformation matrix is $A=\left[\begin{array}{cc}0.89443& 0.44721\\ -0.44721& 0.89443\end{array}\right]$