Ellipse:Finding major and minor axis

1. Mar 21, 2014

utkarshakash

1. The problem statement, all variables and given/known data
The equation $ax^2 + 2hxy+by^2=1$ represents the equation of ellipse if $h^2-ab<0$. When x=Xcosθ-Ysiinθ and y=Xsinθ+Ycosθ, the above equation transforms to
$\dfrac{X^2}{\alpha^2} + \dfrac{Y^2}{\beta ^2} = 1$ where σ and β are real numbers. Then find the product of major and minor axis of this ellipse in terms of h,a,b.

3. The attempt at a solution

I tried transforming the original equation and ended up with this:

$(a \cos ^2 \theta + b \sin ^2 \theta + 2h \cos \theta \sin \theta)X^2$
$+ (a \sin^2 \theta + b \cos ^2 \theta -2h \sin \theta \cos \theta) Y^2$
$+ (-2a \cos \theta \sin \theta +2b \sin \theta \cos \theta +2h \cos^2 \theta - 2h \sin ^2 \theta) = 1$

The product of the major and minor axis will be 4αβ. For finding αβ I tried to find the product of coefficients of X^2 and Y^2. But the expression seems too complicated.

Last edited by a moderator: Mar 21, 2014
2. Mar 21, 2014

Pranav-Arora

Hi utkarshakash!
Erm....where is XY?

Last edited by a moderator: Mar 21, 2014
3. Mar 21, 2014

vanhees71

The idea of the "principal-axis transformation" is to find a orthogonal transformation to diagonalize the representing matrix of the bilinear form on the left-hand side of the equation.

Here you make the ansatz of an arbitrary O(1) transformation. You've already plugged it into your bilinear form. Now you should find $\theta$.

Hint: To that purpose you can drastically simplify the coefficient in front of the "mixed term" $X Y$ by using
$$\sin(2 \theta)=2 \sin \theta \cos \theta, \quad \cos(2 \theta)=\cos^2 \theta-\sin^2 \theta.$$

4. Mar 21, 2014

utkarshakash

I got the correct answer but the calculations were extremely long and tedious. I just want to know if there is some other method which would involve less calculations?

Last edited: Mar 21, 2014
5. Mar 21, 2014

I like Serena

Hi utkarshakash!

I'm not sure how much you know and can work with, so let me just launch into the story and we'll see.

You can write the equation in matrix form as:
$$\begin{pmatrix}x & y\end{pmatrix} \begin{pmatrix}a&h \\ h&b\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} = 1$$
Let's call that matrix $A$. Since the matrix is symmetric, it is diagonalizable with an orthogonal set of eigenvectors.
The diagonalized form corresponds to the transformed equation.

That means that the 2 eigenvalues of the matrix are identical to $1/\alpha^2$ respectively $1/\beta^2$.
Calculate the determinant $\det A=ab-h^2$ to find the product of the eigenvalues.

From there you should be able to deduce the product $2\alpha \cdot 2\beta$...

6. Mar 21, 2014

Staff: Mentor

utkarshakash,
The equation in your attempt in post #1 was way too long, causing the rendered equation to take up too much room. In the future, if you have a very long equation in LaTeX, please split it into multiple lines.

7. Mar 21, 2014

vela

Staff Emeritus
Not really, but the answer to your question really depends on your background in math. The fact that you think this problem was "extremely long and tedious" suggests you don't have a lot of experience, so I'd guess there's not really a method you're familiar with that requires fewer calculations.

This problem, moreover, was essentially was just algebra and trig. As you get more practice, you'll find it easier to do these types of calculations. I would recommend, in fact, that you go back over your work and see if you can streamline the calculations now that you see how things worked out in the end. It will help you develop your intuition on how to group terms, etc.

8. Mar 21, 2014

utkarshakash

I could easily group similar terms(converting trig terms to sin2θ and all that) but I still find it long. I'm only saying this because If I would face this kind of question in an exam, I'm surely going to lose my precious time for just one question.

9. Mar 23, 2014

haruspex

Not sure how you did it since you've baulked at posting all your working, but to add to the above...
The requirement for a zero coefficient on XY gives you an expression for tan(2θ).
On taking the product of the X2 and Y2 coefficients, all of the terms involving θ can be collapsed to sin(4θ) and cos(4θ). Then you can use the usual formulae to express those in terms of tan(2θ).

10. Mar 24, 2014

utkarshakash

I did exactly what you said.