Write the equation into the standard form of the equation of the ellipse: $31x^2+ 10\sqrt{3}xy + 21y^2 − 32x + 32\sqrt{3}y − 80 = 0$.

Question

Write the equation into the standard form of the equation of the ellipse: $$31x^{2}+10\sqrt{3}xy + 21y^{2} − 32x + 32\sqrt{3}y − 80 = 0 $$

I tried but I couldn't solved it. Heres how i tried it. Here's how i tried solving it

$31x^{2}+10\sqrt{3}xy + 21y^{2} − 32x + 32\sqrt{3}y − 80 = 0$

$\Rightarrow(5x)^{2}+6x^{2}+2x.5x\sqrt{3}y+(\sqrt{3}y)^{2}+18y^{2}-32x+32\sqrt{3}y-80=0$

$ \Rightarrow(5x+\sqrt{3}y)^{2}+6x^{2}+18y^{2}-32x+32\sqrt{y}-80=0$

$\Rightarrow (5x+\sqrt{3}y+16)^{2}+6x^{2}+18y^{2}-192x-336=0$

$\Rightarrow (5x+\sqrt{3}y+16)^{2}+6(x^{2}+3y^{2}-32x-56)=0$

$\Rightarrow (5x+\sqrt{3}y+16)^{2}+6\{(x-16)^{2}-1872+18y^{2}\}=0$

Answer 1

If you do the substitution$$\left\{\begin{array}{l}x=\sqrt3X-\frac1{\sqrt3}Y\\y=X+Y\end{array}\right.\tag1$$then$$31 x^2+10 \sqrt{3} x y-32 x+21 y^2+32 \sqrt{3} y-80$$becomes$$144 X^2+\frac{64 Y^2}{3}+\frac{128 Y}{\sqrt{3}}-80=144X^2+\frac{64}{3} \left(Y+\frac{\sqrt{3}}{2}\right)^2-96.\tag2$$It follows from $(1)$ that$$\left\{\begin{array}{l}X=\frac14\left(\sqrt3x+y\right)\\Y=\frac14\left(-\sqrt3x+3y\right)\end{array}\right.$$and so the RHS of $(2)$ is equal to$$144\left(\frac14\left(\sqrt3x+y\right)\right)^2+\frac{64}3\left(\frac14\left(-\sqrt3x+3y\right)+\frac{\sqrt{3}}{2}\right)^2-96.$$Can you take it from here?

Answer 2

A general method is to apply a suitable rotation of angle $\theta$. Let consider the general case: $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ and apply the substitution \begin{align*} x &= x' \cos \theta - y' \sin \theta\\ y &= x' \sin \theta + y' \cos \theta. \end{align*} We will get $$A'x'^2 + B'x'y' + C'y^2 + D'x' + E'y + F' = 0$$ with $B' = B \cos 2\theta - (A - C) \sin 2\theta$. We can then choose a rotation that make $B' = 0$, which $\theta$ then satisfies $$\tan{2\theta} = \frac{B}{A - C}.$$

In your case, we have $\tan 2\theta = \frac{10\sqrt 3}{31 -21} = \sqrt 3$. We can then choose $\theta = \frac{\pi}{6}$ and substitute \begin{align*} x &= x'\cos \frac{\pi}{6} - y' \sin \frac{\pi}{6}\\ y &= x' \sin \frac{\pi}{6} + y' \cos \frac{\pi}{6}. \end{align*}

Answer 3

I am trying to give a visual geometrical solution (including steps of my simple naive approach, which I hope are correct):

First of all, your ellipse looks like this:

Let us rotate this ellipse by $-\frac{\pi}{3}$ using the following transformation:

• $x'=x\cos \left(-\frac{\pi }{3}\right)-y \sin \left(-\frac{\pi }{3}\right)$
• $y'=x \sin \left(-\frac{\pi }{3}\right)+y \cos \left(-\frac{\pi }{3}\right)$

Then we get the following more manageable form:

It has the equation (standard ellipse form):

$$\frac{(x'-2)^2}{3^2}+\frac{(y'-0)^2}{2^2}=1$$

Let us now re-substitute the rotation back (by the angle $\frac{\pi}{3}$) which leads to:

$$ \frac{\left(x\cos\frac{\pi }{3}-y\sin\frac{\pi }{3}-2\right)^2}{3^2}+\frac{\left(x\sin\frac{\pi }{3}+y\cos\frac{\pi}{3}-0\right)^2}{2^2}=1 $$

Answer 4

First, convert the given equation into matrix-vector form.

Define the position vector $r = \begin{bmatrix} x \\ y \end{bmatrix} $

And let

$Q = \begin{bmatrix} 31 && 5 \sqrt{3} \\ 5 \sqrt{3} && 21 \end{bmatrix}$

$b = \begin{bmatrix} -32\\32\sqrt{3} \end{bmatrix} $

$c = -80$

So that the given equation can be written as

$ r^T Q r + b^T r + c = 0 \hspace{24pt} (1) $

First, we'll find the center of the ellipse, this is given by

$\begin{equation} \begin{split} r_0 &= - \frac{1}{2} Q^{-1} b = -\frac{1}{2} \cdot \dfrac{1}{576} \begin{bmatrix} 21 && - 5 \sqrt{3} \\ -5 \sqrt{3} && 31 \end{bmatrix} \begin{bmatrix} -32 \\ 32 \sqrt{3} \end{bmatrix} \\ &= \begin{bmatrix} 1 \\ -\sqrt{3} \end{bmatrix} \end{split} \end{equation}$

Now equation (1) can be re-written as follows:

$ (r - r_0)^T Q (r - r_0) - r_0^T Q r_0 + c = 0 \hspace{24pt} (2) $

We have

$r_0^T Q r_0 = 31 + 21(3) - 2 (5)(3) = 64 $

Hence, equation (2) becomes

$ (r - r_0)^T Q (r - r_0) = 144 \hspace{24pt} (3) $

The next and final step is to diagonalize Q. This has a well-known technique, namely, take

$\theta = \frac{1}{2} \tan^{-1} \dfrac{ 2 Q_{12} }{ Q_{11} - Q_{22} } = \frac{1}{2} \tan^{-1} \dfrac{ 10 \sqrt{3}}{10} = \frac{1}{2} \tan^{-1} \sqrt{3} = \dfrac{\pi}{6}$

Then define the rotation matrix $R$

$R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $

And compute the diagonal matrix $D = \text{diag} \{ D_{11}, D_{22} \} $ as follows

$D_{11} = Q_{11} \cos^2 \theta + Q_{22} \sin^2 \theta + 2 Q_{12} \sin \theta \cos \theta = 31 (3/4) + 21 (1/4) + 2 (5 \sqrt{3})(\sqrt{3}/4) = 36$

$D_{22} = Q_{11} \sin^2 \theta + Q_{22} \cos^2 \theta - 2 Q_{12} \sin \theta \cos \theta = 31(1/4) + 21 (3/4) - 2 (5 \sqrt{3})(\sqrt{3}/4) = 16$

With this, the matrix $Q = R D R^T$ , with $R$ and $D$ as defined above. Substituting this into (3),

$ (r - r_0)^T R D R^T (r - r_0) = 144 \hspace{24pt} (4) $

Divide through by 144, you get,

$ (r - r_0)^T R E R^T (r - r_0) = 1 \hspace{24pt} (5) $

where $E = \frac{1}{144} D = \begin{bmatrix} \frac{1}{4} && 0 \\ 0 && \frac{1}{9} \end{bmatrix} $

Now define the new coordinates of shifted/rotated frame as

$r' = R^T (r- r_0) $ where $r' = \begin{bmatrix} x' \\ y' \end{bmatrix} $ is the position vector with respect to the frame $O'x'y'$ that has its origin at $r_0$ and is rotated with respect to $Oxy$ frame by the angle $\theta$. With this we can now write

$ r'^T E r' = 1$

which translates to,

$\dfrac{x'^2}{4} + \dfrac{y'^2}{9} = 1 $

And this is an ellipse with semi-major axis (along the $y'$ axis) of length $3$ and semi-minor axis (along the $x'$ axis) of length $2$.