Hi guys. Could someone explain to me how the quadric x^2+8xy+7y^2=9 is equivalent to the hyperbola u^2 - v^2/9 =1? Thanks
Something I've just come across while cramming for my linear algebra exam. Going to try and explain the theory first, but once we actually get to working it out it's not that long.
Since the xy term has disappeared we want to orthogonalise the curve, so we will need the quadratic form for the curve. The quadratic form for a curve is given by
&=&\underset{\sim}{x}^{T}A\underset{\sim}{x})
where
is the matrix of the coefficients of the curve, that is the coefficients of powers of
go in the first column, starting with the highest power, then the coefficients of the
terms starting with the highest power go in the second column again starting with the highest power and working downwards.
Now if we try to orthogonalise the matrix by using
where
is the matrix that is formed by taking the nullspace of
corresponding to each eigenvector and making that the columns of
and
is the matrix with the eigenvalues on it's diagonal, and all other entries
. This means that
^{T}D\left(P^{T}\underset{\sim}{x}\right)<br />\\ & =Q\left(\underset{\sim}{x}\right)<br />\\ \text{If }\underset{\sim}{x}=P\underset{\sim}{y}<br />\\ \underset{\sim}{y} & =P^{-1}\underset{\sim}{x}<br />\\ & =P^{T}\underset{\sim}{x}\:\left(\text{as the matrix is a diagonal matrix}\right)<br />\\ \implies Q\left(\underset{\sim}{x}\right) & =\underset{\sim}{y}^{T}D\underset{\sim}{y}<br />\\ & =\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\dots+\lambda_{n}y_{n}^{2}<br />\end{alignedat})
Which means we can take the eigenvalues as the coefficients of the squared terms. Now this will become useful later on.
Any conics can be written in the form
.
Now if
then we can write it in the below form:
.
Now if we divide by
this will give
.
Now we have our quadratic from, we can rewrite the quadratic form in such a way that the cross terms vanish, obtaining a form called standard position. This has the form below:
Where
and
are the eigenvalues of the coefficient matrix
!
Expanding out the above gives
Which we can see that if both the eigenvalues are positive, we will get an ellipse, if one is positive and one is negative we get a hyperbola and if both are negative we get no graph.
So we can take the coefficient matrix
and the eigenvalues will be the coefficients on
and
when we orthogonalise the matrix. (How cool is that!
)
Now to the actual working.
So firstly we need to get the equation into the right form.
Next we find the matrix
which is the coefficient matrix. The coefficients for
will be halfed, as we need to account for if we expanded the matrix back out. In our case this is
Now we need to find the eigenvalues of
, to form the matrix
.
 & =\begin{vmatrix}\lambda-\frac{1}{9} & -\frac{4}{9}<br />\\ -\frac{4}{9} & \lambda-\frac{7}{9}<br />\end{vmatrix}<br />\\ & =\left(\lambda-\frac{1}{9}\right)\left(\lambda-\frac{7}{9}\right)-\frac{16}{81}<br />\\ & =\frac{1}{9}\left(9\lambda^{2}-8\lambda-1\right)<br />\\ & =\frac{1}{9}\left(9\lambda+1\right)\left(\lambda-1\right)<br />\\ & =0<br />\\ \implies\lambda_{1} & =1<br />\\ \lambda_{2} & =-\frac{1}{9}<br />\\ u^{2}-\frac{1}{9}v^{2} & =1<br />\end{alignedat})
Which is what we wanted to get to. Hope that helps (and makes sense)
....now back to cramming
(banning myself from AN for a few days to get work done).
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