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What are major & minor axes of the ellipse: $10x^2+14xy+10y^2-7=0$ ?

My trial: from given equation: $10x^2+14xy+10y^2-7=0$

$$10x^2+14xy+10y^2=7$$ $$\frac{x^2}{7/10}+\frac{xy}{7/14}+\frac{y^2}{7/10}=1$$ I know the standard form of ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ But the term of $xy$ is a bottleneck.

From here I can't proceed. Can somebody please help me solve this problem?

Thank you.

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    $\begingroup$ For a second order polynomial write it as a quadratic form with matrices and do an eigenvalue decomposition. $\endgroup$ – mathreadler Jan 22 '18 at 7:56
  • $\begingroup$ If there is any analytic method to solve it please give more hint or explain it $\endgroup$ – jeanne clement Jan 22 '18 at 7:58
  • $\begingroup$ There are many ways to solve this, Jeanne. But to make it smooth you should tell us a few bits about your background. Have you taken linear algebra? The reason for asking that is that a very easy method relies on the theory of eigenvalues of symmetric matrices. But if that is all Greek to you, then we need to do something else. Another possibility (taken by the answerer) depends on you being familiar with coordinate rotations. Have you covered that? Anyway, you identified the problem: it is the mixed term $xy$. Getting rid of that with a change of coordinates (one way or another) is the key. $\endgroup$ – Jyrki Lahtonen Jan 22 '18 at 8:24
  • $\begingroup$ Here you can also observe that because the coefficients of $x^2$ and $y^2$ are equal, the ellipse is tilted 45 degrees. You can try and write the equation in terms of $x'=(x+y)$ and $y'=(x-y)$ (you may or may not want to divide those by $\sqrt2$, at some point you need to adjust for that). $\endgroup$ – Jyrki Lahtonen Jan 22 '18 at 8:28
  • $\begingroup$ I have time to give a better explanation in the afternoon, but I suspect other people will be faster. $\endgroup$ – mathreadler Jan 22 '18 at 9:48
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Rewriting the equation in polar coordinates, with $x=r\cos\theta$ and $y=r\sin\theta$, we get

$$r^2(10\cos^2\theta+14\cos\theta\sin\theta+10\sin^2\theta)-7=0$$

Using the trig identities $\sin^2\theta+\cos^2\theta=1$ and $2\sin\theta\cos\theta=\sin2\theta$, we find this simplifies to

$$r^2(10+7\sin2\theta)-7=0$$

or

$$r=\sqrt{7\over10+7\sin2\theta}$$

The largest value of $r$ occurs when $\sin2\theta=-1$ and the smallest when $\sin2\theta=1$. Thus the major axis is $2\sqrt{7/3}$ and the minor axis is $2\sqrt{7/17}$.

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  • $\begingroup$ if the $coeff(x^2)\neq coeff(y^2)$, this method will have an extra $cos^2\theta$ or $\sin^2\theta$ left over, which can be converted to $\cos2\theta$, and then the final expression for $r$ that we get will contain $a\sin(2\theta)+b\cos(2\theta)$. Apart from that, the method remains same. Is my analysis correct? $\endgroup$ – Gaurang Tandon Jan 24 '18 at 13:29
  • $\begingroup$ Also, how will you get this method to work in case of the general equation $ax^2+2hxy+by^2+2gx+2fy+c=0$? There will be an extra term of $\sin\theta$ or $\cos\theta$ left over and I am not sure how to deal with it :/ $\endgroup$ – Gaurang Tandon Jan 24 '18 at 13:29
  • $\begingroup$ @GaurangTandon, you are quite right, my approach works as easily as it does only because of the simple form of the equation in question. It's just one arrow to consider in a quiver of techniques. $\endgroup$ – Barry Cipra Jan 24 '18 at 15:34
  • $\begingroup$ It's the quickest arrow for this question, I learnt something new, thanks! :D $\endgroup$ – Gaurang Tandon Jan 24 '18 at 15:35
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The ellipse we have is in the form $S(x, y) = ax^2 + by^2 + 2hxy + c$. To remove the $xy$ term we rotate the ellipse by the angle $\tan (2\theta) = \dfrac{2h}{a - b}$. Which is same as rotating coordinate system by $-\theta$.

So if $(X', Y')$ is coordinates in new coordinate system then we can use

$${x \choose y} = \left[\begin{matrix}\cos\theta & -\sin \theta \\ \sin \theta& \cos \theta \end{matrix}\right] {X^\prime \choose Y^\prime}$$

to find relation between old coordinate system $(x,y)$ and new coordinate system $(X', Y')$.

Plugging $ \theta = -\pi/4$

$${x \choose y} = \dfrac{1}{\sqrt{2}}\left[\begin{matrix}1 & 1 \\ -1& 1 \end{matrix}\right] {X^\prime \choose Y^\prime} = \dfrac{1}{\sqrt{2}}{X' +Y'\choose Y'-X'}$$

So, $$10x^2+14xy+10y^2-7= 3 X'^2 + 17 Y'^2 - 7 \implies \dfrac{X'^2}{7/3} + \dfrac{Y'^2}{7/17} = 1$$

The major axis is $2\sqrt{\dfrac73}$ and minor axis is $2\sqrt{\dfrac7{17}}$.

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  • $\begingroup$ how do you know ellipse should be rotated by $\pi/4$? $\endgroup$ – jeanne clement Jan 22 '18 at 8:20
  • $\begingroup$ Jeanne I used $\tan (2\theta) = \dfrac{2h}{a - b}$ to get theta. Sorry I edited the answer properly, I made a small mistake in original answer. The rotation angle is $-\pi/4$ not $\pi/4$. The ellipse is already tilted by $\pi/4$ so by rotating the plane we negate that tilt. $\endgroup$ – user8277998 Jan 22 '18 at 8:21
  • $\begingroup$ How to know that direction of rotation of ellipse either $+\theta$ or $-\theta$ is there any criteria to know direction of rotation? As you said 'The rotation angle is $-\pi/4$ not $\pi/4$' how did you know that? If I take $\pi/4$ as rotation angle then will it be wrong? $\endgroup$ – jeanne clement Jan 22 '18 at 9:41
  • $\begingroup$ @jeanneclement I don't. The ellipse is rotated by some angle $\theta$, to negate that rotation we just rotate plane in opposite direction i.e $-\theta$. The ellipse in your question is rotated by $\pi/4$ so we rotate the plane by $-\pi/4$. $\endgroup$ – user8277998 Jan 22 '18 at 9:44
  • $\begingroup$ @jeanneclement Does desmos.com/calculator/wdpv7j5p6i help ? $\endgroup$ – user8277998 Jan 22 '18 at 9:46
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The set $E\subset{\mathbb R}^2$ defined by the given equation is symmetric with respect to the lines $y=\pm x$, since $$(x,y)\in E\Leftrightarrow (y,x)\in E,\qquad (x,y)\in E\Leftrightarrow(-y,-x)\in E\ .$$ It follows that these two lines are the principal axes of $E$. Putting $y=x$ in the equation gives $34x^2=7$, or $x=\pm\sqrt{7/34}$, and putting $y=-x$ gives $6x^2=7$, or $x=\pm\sqrt{7/6}$. It follows that length of the mayor semiaxis is $a=\sqrt{7/3}$, and the length of the minor semiaxis is $b=\sqrt{7/17}$.

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Yet another method is with matrices and linear algebra. If we stuff $(x,y,1)^T$ into a vector, then the matrix

$${\bf M}=\left[\begin{array}{ccc}10&7&0\\7&10&0\\0&0&-7\end{array}\right]$$

Can be used to express the ellipse as a scalar product with matrix multiplication:

$$\left[\begin{array}{ccc}x&y&1\end{array}\right] \left[\begin{array}{ccc}10&7&0\\7&10&0\\0&0&-7\end{array}\right] \left[\begin{array}{c}x\\y\\1\end{array}\right]$$

A linear coordinate change can be expressed like this:

$$\left[\begin{array}{c}x_1\\y_1\\1\end{array}\right]= {\bf T}_{0\to 1}\left[\begin{array}{c}x_0\\y_0\\1\end{array}\right]$$

By making an eigenvalue decomposition of $\bf M$ we can find the coordinate system to transform to:

$${\bf M} = \left[\begin{array}{ccc}0&-1/\sqrt 2&1/\sqrt 2\\0&1/\sqrt 2&1/\sqrt 2\\1&0&0\end{array}\right]\left[\begin{array}{ccc}-7&0&0\\0&3&0\\0&0&17\end{array}\right]\left[\begin{array}{ccc}0&-1/\sqrt 2&1/\sqrt 2\\0&1/\sqrt 2&1/\sqrt 2\\1&0&0\end{array}\right]^{-1}$$

Here the coefficients of the new expression appears in the diagonal: $-7,3, 17$. The rest of the steps are similar to @user45914123 answer.


The main benefit of this method is that we can bake translation into the base change. For example if you had $X^2+2X+1+Y^2=R^2$, you would not know what to do with the $2X$ if you could only do linear transformations on the $[X,Y]^T$ vector.

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