Area of an ellipse.

Question

I need to find the area of the image of a circle centred at the origin with radius 3 under the transformation:

$ \begin{pmatrix} 3 & 0\\ 0 & \frac{1}{3} \end{pmatrix} $

The image is the ellipse $ \frac{x^2}{81}+y^2=1$. It would appear that it has the same area as the original circle i.e. $9\pi$. Is this because the matrix has some special property such as being its own inverse?

It's because the matrix has the property that its determinant is 1. By the way your equation doesn't match the matrix, perhaps the equation should be $x^2/9 + 9y^2=1$. — KalEl, May 26 '15 at 11:01
Think geometrically. You've stretched the $x$ axis and shrunk the $y$ axis. — Ethan Bolker, May 26 '15 at 11:02
Also by the way so you are clear - the matrix is not its own inverse. — KalEl, May 26 '15 at 11:07
Yes, thank you for pointing out that it's not its own inverse. That was a big mistake. — bibo_extreme, May 26 '15 at 11:09

score 2 · Accepted Answer · answered May 26 '15 at 11:04

2

Yes, this matrix has a special property, namely its determinant is 1.

answered May 26 '15 at 11:04

citronrose

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Is there a particular name or result for your answer? – bibo_extreme May 26 '15 at 11:10
May this question will help: http://math.stackexchange.com/q/668/162363 – citronrose May 26 '15 at 11:21

score 0 · Answer 2 · answered May 26 '15 at 11:39

0

It is a known formula that the area enclosed in the ellipse with semi-axes $a$ and $b$ is $\pi ab$, as may be seen from the orthogonal affinity that transforms a circle into an ellipse.

answered May 26 '15 at 11:39

Bernard

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Area of an ellipse.

2 Answers2