Finding the affine transformation that will change a given ellipse into the unit circle $x^2$ + $y^2$ =1.

Question

We are given that ellipse $E$ is given by $x^2+4y^2-2x+16y+1=0$ and we are asked to find $t_2 \in A(2)$ such that $t_2(E)$ is the unit circle.

Answer 1

Transformation is - tranlsation + scaling $$ x^2+4y^2-2x+16y+1=x^2-2x+1+4(y^2+4y+4)-16=(x-1)^2+4(y+2)^2-16=0 \\ \frac{(x-1)^2}{16}+\frac{(y+2)^2}4=1 $$ So if you change coordinates $$ \xi = \frac{x-1}4 \\ \eta= \frac{y+2}2 $$ your ellipse will become $\xi^2+\eta^2=1$

Update

Transformation itself can be represented in vector form $\mathbf x' = A \mathbf x + \mathbf b$ $$ \left [ \begin{array}{c} \xi \\ \eta \end{array}\right ] = f \left ( \left [ \begin{array}{c} x \\ y \end{array}\right ]\right ) = \left [ \begin{array}{c} \frac x4 - \frac 14 \\ \frac y2 + 1 \end{array}\right ] = \left [ \begin{array}{cc} \frac 14 & 0 \\ 0 & \frac 12 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array}\right ] + \left [ \begin{array}{c} -\frac 14 \\ 1 \end{array}\right ] $$