In the euclidean space $E 2$ consider the ellipse $C$ with foci $F_1 = (1, 7)$ and $F_2 = (7, 7)$,
passing through the point $(4, 3)$. (i) Find a cartesian equation of $C$. (ii) Find a cartesian equation of the canonical form $C'$ of $C$.
(iii) Find an isometry which maps $C$ to $C$ ' .
Solution
(i)
$16x^2-128x+25y^2-350y+1081=0$
(ii)
$\frac{(x-4)^2}{25} +\frac{(y-7)^2}{16}=1$
(iii)
This part of question is a little bit ambiguous,but I assume that I am asked to find the matrix representing isometry with respect to the canonical basis.
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If i take two points on ellipse $(4,11)$ and $(9,7)$ i would have $(4,11) \rightarrow (4,11)$ and $(9,7)\rightarrow(9,7)$.Then, $11\times (9,7)-7\times(4,11)\rightarrow 11\times(9,7)-7\times(4,11)$ or $(71,0)\rightarrow(71,0)$ which means $(1,0)\rightarrow(1,0)$.For the other vector of canonical basis i have $(0,1)\rightarrow(0,1)$.Would i get the correct result for (iii) if a transposed two vectors and wrote it as matrix?
There’s nothing ambiguous about the question, although there is more than one possible answer to it.
Construct the isometry in two stages. First, find the translation that takes the center of the ellipse to the origin. Then, find a rotation or reflection that aligns the major axis of the ellipse with the $x$-axis. In this case, the original axis is already parallel to the $x$-axis, so the simplest thing to do is nothing—a pure translation works.
Note, though, that since any isometry that solves this problem involves a translation, it can’t be represented as a matrix unless you use homogeneous coordinates, for which you’d have a $3\times3$ matrix. You always decompose an affine transformation into the composition of a linear transformation and a translation, though, and then represent the linear part by a $2\times2$ matrix, which for the above isometry is just the identity.