Question

(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

Alternate coordinate systems

(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 by the two distinct bases are described as follows, respectively: $$2(x_1)^2 -4(x_1)(y_1) + 5(y_1)^2 - 36 = 0$$ and $$(x_2)^2 + 6(y_2)^2 - 36 = 0.$$ Find the transformation matrix between these two coordinate systems: $$(x_1, y_1)$$ and $$(x_2, y_2)$$.

2021-02-01

Let the transformation matrix be $$A = \begin{bmatrix}P & Q \\ R & S \end{bmatrix}$$

$$A = \begin{bmatrix} X1 \\ Y1 \end{bmatrix} = \begin{bmatrix}P & Q \\ R & S \end{bmatrix} \begin{bmatrix} X1 \\ Y1 \end{bmatrix} = \begin{bmatrix}PX1 + QY1 \\ RX1 + SY1 \end{bmatrix} = \begin{bmatrix} X2 \\ Y2 \end{bmatrix} \ X2 = [PX1 + QY1] \ Y2 = [RX1 + SY1]$$

Putting in thhe given EQN. $$[PX1 + QY1]^2 + 6[RX1 + SY1]^2 - 36 = 2 X1^2 - 4 X1Y1 + 5 Y1^2 - 36 \ P2 + 6R2 = 2 \ Q2 + 6S2 = 5 \ 2PQ + 12RS = -4......PQ + 6RS = -2$$ 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 $$\begin{bmatrix}COS(T) & SIN(T) \\-SIN(T) & COS(T) \end{bmatrix} \begin{bmatrix} X1 \\ Y1 \end{bmatrix} = \begin{bmatrix} X2 \\ Y2 \end{bmatrix}$$

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

Where $$\displaystyle{H}={C}{O}{E}{F}{F}{I}{C}{I}{E}{N}{T}\ {O}{F}\ {X}{1}{Y}{1}=-{4}$$

$$A = COEFFICIENT\ OF\ X1^2 = 2$$

$$B = COEFFICIENT\ OF$$ $$Y1^2 = 5$$

Solving we get $$\displaystyle{P}={0.89443}\ldots\ldots{Q}={0.44721}\ldots\ldots{R}=--{0.44721}\ldots\ldots\ldots{S}={0.89443}$$

Hence the transformation matrix is $$A = \begin{bmatrix} 0.89443 & 0.44721 \\ -0.44721 & 0.89443 \end{bmatrix}$$