- Thread starter
- Banned
- #1

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?

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?

Last edited: