How to find major & minor axes of the ellipse $10x^2+14xy+10y^2-7=0$?

Question

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.

Answer 1

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 $2\sqrt{\dfrac7{17}}$.

Answer 2

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 $2\sqrt{7/17}$.

Answer 3

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 $b=\sqrt{7/17}$.

Answer 4

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.

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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.