1
$\begingroup$

The principal axes of the ellipse given by the equation $ 9x^2+6y^2-4xy=4$ are along the directions of the vectors

a) $2i+3j$ and $3i-2j$ b) $2i+j$ and $i-2j$ c) $3i+2j$ and $2i-3j$ d) $i+2j$ and $2i-j$ e) $i+j$ and $i-j$

Seems trivial however I do not have much idea on how to find those vectors.

$\endgroup$

2 Answers 2

3
$\begingroup$

Notice that

$$ 9x^2+6y^2-4xy = \begin{pmatrix} x& y \end{pmatrix} \begin{pmatrix} 9 & \frac{-4}{2} \\ \frac{-4}{2} & 6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$ We'll define $\vec{x} = \begin{pmatrix} x \\ y \end{pmatrix} $. Now: $$ 9x^2+6y^2-4xy = \vec{x}^T \begin{pmatrix} 9&-2\\ -2&6 \end{pmatrix} \vec{x} $$ The matrix $\begin{pmatrix} 9&-2\\ -2&6 \end{pmatrix}$ can be diagonalized, which we prefer because then in the new basis there will be no xy term, meaning the new basis will give us the principal axes of the ellipse. Let's find the eigenvalues: $$ (\lambda-9)(\lambda - 6) - 4 = \lambda^2 - 15\lambda + 50 = 0$$ $$ \lambda = 5 ,10 $$ For $\lambda = 5$ we'll find the eigen vector: $$\begin{pmatrix} 4&-2 \\ -2 & 1 \end{pmatrix} v = 0$$ It is easy to see that $v=\begin{pmatrix} 1\\2 \end{pmatrix} $ solves the equation. The original matrix is a symmetric matrix meaning the other eigen vector will be orthogonal to the one we just found meaning it will be $\begin{pmatrix} 2\\-1 \end{pmatrix} $ So the answer is option (d).

$\endgroup$
6
  • $\begingroup$ Here is a nice graph from wolfram alpha: wolframalpha.com/input/… $\endgroup$ May 2, 2017 at 13:27
  • $\begingroup$ Though I understood the solution I did not get how you inferred that you have to diagonalize to get a new basis which is the principal axes. $\endgroup$
    – rahul rj
    May 2, 2017 at 14:06
  • $\begingroup$ Which part would you want me to explain? Do you want me to prove that in the principal axes base the matrix is a diagonal matrix? $\endgroup$ May 2, 2017 at 14:16
  • $\begingroup$ That would be helpful to me. If you've got a link to the relevant topic that deals with problems such as this that would do well too. $\endgroup$
    – rahul rj
    May 2, 2017 at 14:24
  • $\begingroup$ Yeah, sure, I hope this is helpful: maths.qmul.ac.uk/~twm/MTH6140/la26.pdf $\endgroup$ May 2, 2017 at 14:35
1
$\begingroup$

Consider the straight line through the origin

$$y=mx$$ and find its intersections with the ellipse by solving

$$9x^2+6m^2x^2-4mx^2=4,$$

or

$$x^2=\frac4{6m^2-4m+9}.$$

The squared distance from the origin to an intersection is

$$x^2+y^2=x^2(1+m^2)=\frac{4(m^2+1)}{6m^2-4m+9}.$$

By canceling the derivative on $m$, we find the extrema of this function at $$m=2,\\m=-\frac12,$$ corresponding to the direction vectors

$$(1,2),\\(2,-1).$$

The corresponding squared distances are

$$\frac45,\frac25,$$ telling you which are the short and long axis.

$\endgroup$
4
  • 1
    $\begingroup$ I love the fact that both of us solved the same question with completely different mathematical tools, have an upvote! $\endgroup$ May 2, 2017 at 13:38
  • $\begingroup$ @Yves Pardon me if it sounds silly, how did you find m=2? you take the derivative of what exactly? $\endgroup$
    – rahul rj
    May 2, 2017 at 14:00
  • $\begingroup$ @rahulrj of the squared distance. $\endgroup$
    – user65203
    May 2, 2017 at 14:21
  • $\begingroup$ Another method would be to take $x=r\cos\theta$ and $y=r\sin\theta$. $\endgroup$
    – Sawarnik
    May 3, 2017 at 7:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .