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I'm trying to find an affine transformation that maps the unit circle to an ellipse centered at $(1,3)$ such that points P$(-3,-1)$ and R$(5,7)$ are at the greatest distance from the centre of the ellipse along the major axis while points Q$(0,4)$ and S$(2,2)$ are at the greatest distance from the centre of the ellipse along the minor axis.

I think we need to rotate the unit disc followed by scaling and then another rotation. But I don't know how to find the matrix that does that. Ultimately, I think there will be translation by $(1,3)^T.$

Any help appreciated.

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1 Answer 1

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Hints :

  • $(5,7), (1,3), (-3,-1)$ lie on line $y=mx+c$
  • $(0,4), (1,3), (2,2)$ lie on line $y=\ldots$
  • Conclude this is rotation with $\theta = \tan^{-1} m$
  • Use rotation matrix $$ \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix} $$
  • Find lengths of axes of ellipse. Hence find scaling matrix. $$ \begin{bmatrix} a & 0 \\ 0 & b \\ \end{bmatrix} $$
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  • $\begingroup$ So, am I right in saying that we have to rotate the unit disc by the rotation matrix where $\theta = \dfrac{\pi}{4}$ followed by scaling matrix where $a$ and $b$ are $4$ and $1$ respectively and then translation by $(1,3)^T$ ? So do we only need one rotation ? $\endgroup$
    – user715112
    Commented Nov 17, 2020 at 15:31
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    $\begingroup$ @Rhombus Almost! It's $4\sqrt{2}$, $\sqrt{2}$. $\endgroup$
    – cosmo5
    Commented Nov 17, 2020 at 15:33
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    $\begingroup$ Yes. Only one rotation is required. $\endgroup$
    – cosmo5
    Commented Nov 17, 2020 at 15:36

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