0

I am trying to solve the image of a tilted ellipse $x^2+3y^2-3xy-8x+10y+1=0$ under the linear transformation represented by the matrix

$\begin{bmatrix}-1&1\\1&3\end{bmatrix}$

Since there are six terms in the given equation, I am thinking of finding 6 points on the curve and find their respective image using the given matrix and find the values of the numerical coefficients of the following equation.

$ax^2+by^2+cxy+dx+ey+f=0$

I would like to kindly ask if my solution is possible and if there are other simpler ways to find the image of the linear transformation.

Thanks in advance.

PRD
  • 611
  • 1
    I suggest to take a look at Roberts answer here- https://math.stackexchange.com/questions/3200569/geometrically-explained-why-do-linear-transformations-take-a-circle-to-an-ellip, and just make your changes in the circle equation (multiply it by the respective matrices to make it an ellipse) – Or Shahar Nov 15 '22 at 07:10
  • @OrShahar, Thank you for sharing this past questions and Robert's answer. I got the idea but could you give some hint on how I can use the given to transform it into a matrix? – PRD Nov 15 '22 at 07:22

1 Answers1

1

Write the ellipse as $$[x,y,1]\begin{bmatrix}1&-3/2&-4\\-3/2&3&5\\-4&5&1 \end{bmatrix}\begin{bmatrix}x\\y\\1 \end{bmatrix}=0.$$Make the linear transformation $$\begin{bmatrix}x^{\prime}\\y^{\prime} \\1\end{bmatrix}=\begin{bmatrix}-1&1&0\\1&3&0\\0&0&1 \end{bmatrix}\begin{bmatrix} x\\y\\1\end{bmatrix}.$$Note that the inverse of $$\begin{bmatrix}-1&1&0\\1&3&0\\0&0&1 \end{bmatrix}$$ is $$\begin{bmatrix}-3/4&1/4&0\\1/4&1/4&0\\0&0&1 \end{bmatrix}$$

P. Lawrence
  • 5,674
  • Thank you! May I kindly ask, why do we need the inverse if what we need is to find $\begin{bmatrix}x^{\prime}\y^{\prime} \1\end{bmatrix}$? – PRD Nov 15 '22 at 23:44
  • Do I need to express $[x,y,1]$ and $\begin{bmatrix}x\y \1\end{bmatrix}$ in terms of $\begin{bmatrix}x^{\prime}\y^{\prime} \1\end{bmatrix}$? – PRD Nov 15 '22 at 23:53
  • Please correct me if I'm wrong.

    $\begin{bmatrix}x^{\prime}\y^{\prime} \1\end{bmatrix} = \begin{bmatrix}-1&1&0\1&3&0\0&0&1 \end{bmatrix}\begin{bmatrix}x\y \1\end{bmatrix}$. Hence,

    $\begin{bmatrix}x\y \1\end{bmatrix} = \frac{1}{4} \begin{bmatrix}-3&1&0\1&1&0\0&0&4 \end{bmatrix}\begin{bmatrix}x^{\prime}\y^{\prime} \1\end{bmatrix}$

    $[x,y,1] = \frac{1}{4} [x^{\prime},y^{\prime},1]\begin{bmatrix}-3&1&0\1&1&0\0&0&4 \end{bmatrix}$

     – PRD Nov 16 '22 at 00:21
  • Based on the first matrix equation, the answer must be

    $\frac{1}{16} [x^{\prime},y^{\prime},1]\begin{bmatrix}-3&1&0\1&1&0\0&0&4 \end{bmatrix}\begin{bmatrix}1&-3/2&-4\-3/2&3&5\-4&5&1 \end{bmatrix}\begin{bmatrix}-3&1&0\1&1&0\0&0&4 \end{bmatrix}\begin{bmatrix}x^{\prime}\y^{\prime} \1\end{bmatrix}=0$

     – PRD Nov 16 '22 at 00:23
  • 1
    @PRD: Suppose we have a conic $$[x y 1]M\begin{bmatrix}x\y\1\end{bmatrix}=0.$$Suppose also that we have an invertible transformation $$\begin{bmatrix}x^{\prime}\y^{\prime}\1\end{bmatrix}=P\begin{bmatrix}x\y\1\end{bmatrix}.$$ Suppose that $(a,b)$ is a point on the conic and that $(\alpha,\beta)$ is its image under the transformation ,so $$\begin{bmatrix}\alpha\\beta\1\end{bmatrix}=P\begin{bmatrix}a\b\1\end{bmatrix}$$ and $$[a b 1]M\begin{bmatrix}a\b\1\end{bmatrix}=0.$$ contnued innext comment – P. Lawrence Nov 16 '22 at 22:33
  • 1
    Since $$\begin{bmatrix}a\b\1\end{bmatrix}=P^{-1}\begin{bmatrix}\alpha\\beta\1\end{bmatrix},$$ we have $$\alpha \beta 1^TMP^{-1}\begin{bmatrix}\alpha\\beta\1\end{bmatrix}=0.$$ Thus the transformed point $(\alpha,\beta)$ lies on the conic $$x y 1^TMP^{-1}\begin{bmatrix}x\y\1\end{bmatrix}=0.$$ – P. Lawrence Nov 16 '22 at 22:36