Translation of x^2+xy+y^2+2x-3y+5
- Thread starter Goistein
- Start date
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,976
Given the general conic \(\displaystyle Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), in order to determine the angle of rotation to eliminate the \(\displaystyle Bxy\) term solve this:
\(\displaystyle 2\phi = \arctan \left( {\frac{B}{{\left( {A - C} \right)}}} \right)\).
Then: \(\displaystyle \left[ {\begin{array}{c}
x \\
y \\
\end{array}} \right] = \left[ {\begin{array}{lr}
{\cos (\phi )} & { - \sin (\phi )} \\
{\sin (\phi )} & {\cos (\phi )} \\
\end{array}} \right]\left[ {\begin{array}{c}
{x'} \\
{y'} \\
\end{array}} \right]\)
\(\displaystyle 2\phi = \arctan \left( {\frac{B}{{\left( {A - C} \right)}}} \right)\).
Then: \(\displaystyle \left[ {\begin{array}{c}
x \\
y \\
\end{array}} \right] = \left[ {\begin{array}{lr}
{\cos (\phi )} & { - \sin (\phi )} \\
{\sin (\phi )} & {\cos (\phi )} \\
\end{array}} \right]\left[ {\begin{array}{c}
{x'} \\
{y'} \\
\end{array}} \right]\)
Actually, here's an example of how the question is supposed to be done (I just don't understand the solution):
We want to sketch the quadratic curve
4x2 - 4xy + 7y2 +12x + 6y - 9 = 0.
Since
A = 4, B = -2 and C = 7,
we get [[4, -2][-2, 7]]=24>7
Thus, the curve is going to be 'elliptic'. The origin shift to ( x0, y0 )is given by the solution of the
equations
4x0 - 2y0 + 6 = 0
-2x0 + 7y0+ 3 = 0.
This gives x0 = -2, y0 = -1. The quadratic form of the transformed curves is given by
4x2 - 4xy + 7y2 - 24 = 0.
Equivalently [[x, y]][[4, -2][-2, 7]][[x][y]]=24
Thus A=[[4, -2][-2, 7]]
Its eigenvalues are 8 and 3 with respective eigenvectors
[[1][-2]] and [[2][1]]
Thus, he quadratic curve is
- 8( x/ )2 + 3( y' )2 = 24.
I'm not sure how the equations in red were obtained and is the 7 in orange supposed to be the "C" number? Oh, and since the matrices are kinda sloppy, the site is:
http://www.mathresource.iitb.ac.in/line ... r11.3.html
It's 11.3.6 Example
We want to sketch the quadratic curve
4x2 - 4xy + 7y2 +12x + 6y - 9 = 0.
Since
A = 4, B = -2 and C = 7,
we get [[4, -2][-2, 7]]=24>7
Thus, the curve is going to be 'elliptic'. The origin shift to ( x0, y0 )is given by the solution of the
equations
4x0 - 2y0 + 6 = 0
-2x0 + 7y0+ 3 = 0.
This gives x0 = -2, y0 = -1. The quadratic form of the transformed curves is given by
4x2 - 4xy + 7y2 - 24 = 0.
Equivalently [[x, y]][[4, -2][-2, 7]][[x][y]]=24
Thus A=[[4, -2][-2, 7]]
Its eigenvalues are 8 and 3 with respective eigenvectors
[[1][-2]] and [[2][1]]
Thus, he quadratic curve is
- 8( x/ )2 + 3( y' )2 = 24.
I'm not sure how the equations in red were obtained and is the 7 in orange supposed to be the "C" number? Oh, and since the matrices are kinda sloppy, the site is:
http://www.mathresource.iitb.ac.in/line ... r11.3.html
It's 11.3.6 Example