Find the standard form of conic

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SUMMARY

The standard form of the conic represented by the equation $7x^2 + 48xy - 7y^2 + 20x - 110y - 50 = 0$ is derived as $(y')^2 - (x')^2 = 1$, indicating it is a hyperbola. The transformation to standard form involves using a matrix of coefficients with eigenvalues of +25 and -25, and eigenvectors (4,3) and (3,-4). The transformation is defined as f(P) = 0.2$\begin{bmatrix}4&3\\3&-4\end{bmatrix}$P - $\begin{bmatrix}1\\2\end{bmatrix}$. A new orthonormal basis $B = \{\frac{1}{5}(4,3), \frac{1}{5}(-3,4)\}$ is utilized for rotation, preserving orientation.

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Poirot1
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Find the standard form of $7x^2+48xy-7y^2+20x-110y-50=0$ and find euclidean transformation taking it to standard form.

Answer: matrix of coefficents has eigenvalues + or -25 with eigenvectors (4,3) and (3,-4) respectively. set x'=1/5(4x+3y+a) and y'=1/5(3x-4+b). then comparing the terms of
$25(x')^2-25(y')^2$ and the equation gives a = -5 and b=-10 so standard form is

$(y')^2-(x')^2=1$. This is a hyperbola, not orientated in the usual way (as I will send y to y' and x to x') but still a hyperbola.

Transformation is f(P)=0.2$\begin{bmatrix}4&3\\3&-4\end{bmatrix}$P-$\begin{bmatrix}1\\2\end{bmatrix}$

Ok?
 
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Poirot said:
Ok?

Right. Using as a new orthonormal basis $B=\{\frac{1}{5}(4,3),\frac{1}{5}(-3,4)\}$ instead of $B'=\{\frac{1}{5}(4,3),\frac{1}{5}(3,-4)\}$ we get a rotation, which keeps the orientation.
 

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