Ellipse $3x^2-x+6xy-3y+5y^2=0$: what are the semi-major and semi-minor axes, displacement of centre, and angle of incline?

Question

Given the ellipse $$3x^2-x+6xy-3y+5y^2=0$$ find the following:

• semi-major axis, $a$
• semi-minor axis, $b$
• displacement of centre from origin (or coordinates of centre of ellipse $(h,k)$)
• angle of incline $\theta$ (or $\tan\theta$).

This can of course be done systematically by equating coefficients with

$$\frac{((x-h)\cos\theta-(y-k)\sin\theta)^2}{a^2}+\frac{((x-h)\sin\theta+(y-k)\cos\theta)^2}{b^2}=1$$

which is derived by applying the rotation and translation parameters to the standard ellipse equation. Perhaps someone could complete the solution?

Also, can it be solved in any other way, and in particular, without using trigonometric ratios?

(NB - this problem came about whilst trying to solve another problem on MSE here How to Solve this Arithmetic Progression Question? which gave rise to the Diophantine Equation above; assuming instead that $x,y$ can take on non-integer values the equation describes a rotated and translated ellipse)

Answer 1

To find the center, translate until the first degree terms vanish.

$$3(x+u)^2-(x+u)+6(x+u)(y+v)-3(y+v)+5(y+v)^2\to\\ 6u-1+6v=0,\\ 6u-3+10v=0.$$ The solution is $(-\frac13,\frac12)$, and the reduced equation $$3x^2+6xy+5y^2-\frac7{12}=0.$$

Then you need to diagonalize the quadratic coefficient matrix $$\begin{pmatrix} 3 & 3 \\ 3 & 5 \end{pmatrix},$$ with characteristic equation $$\lambda^2-8\lambda+6=0,$$ giving the Eigenvalues $4\pm\sqrt{10}.$

After diagonalization, $$(4-\sqrt{10})u^2+(4+\sqrt{10})v^2=\frac7{12},$$ giving the half-axis lengths $$\sqrt{\frac7{12(4-\sqrt{10})}}\text{, and }\sqrt{\frac7{12(4+\sqrt{10})}}.$$ The direction of the long axis is given by some Eigenvector associated to the small Eigenvalue, such as $(-3, \sqrt{10}-1)$, so the angle $$-\arctan\frac{\sqrt{10}-1}3.$$

Answer 2

Whilst acknowledging the very nice solution by Yves Daoust, here's an alternative solution derived earlier using only basic calculus and symmetry of an ellipse about its centre. This is posted for general information only.

1. Centre of Ellipse

Equation of ellipse: $$3x^2-x+6xy-3y+5y^2=0$$

Differentiating w.r.t $x$: $$6x-1+6x\frac{dy}{dx}+6y-3y\frac{dy}{dx}+10y\frac{dy}{dx}=0$$

When $\dfrac{dy}{dx}=0$, $6x+10y=3$.

Substituting into the original equation gives $24x^2+16x-9=0$.

Solving gives $$(x,y)=\left(-\frac13\pm\frac{\sqrt{70}}2,\frac12\mp\frac{70}2\right)$$ which are the coordinates of the highest and lowest points of the ellipse. Since both these points are symmetrical about the point $$\left(-\frac13,\frac12\right)\qquad \blacksquare$$ this must be the centre of the ellipse, given that the ellipse is symmetrical about its centrepoint.

Translate the ellipse such that the centre is at the origin, by substituting $(x,y)$ in the original equation with $(x+h,y+k)$, where $(h,k)=(-\frac13,\frac12)$, which gives the translated ellipse equation as: $$3x^2+6xy+5y^2-\frac{7}{12}=0$$

2. Incline of Ellipse

Consider the intersection of the translated ellipse with the line $y=mx$. The incline of (the major/minor axis of) the ellipse is the value of $m$ when the distance of the intersection points from the origin is at a maximum/minimum.

Substituting $y=mx$ into the original equation gives: $$(3+6m+5m^2)x^2=\frac7{12}$$ Let $l$ be the distance of the intersection point from the origin. $$l^2=x^2+y^2=x^2(1+m^2)=\frac7{12}\underbrace{\left(\frac{1+m^2}{3+6m+5m^2}\right)}_{f(m)}$$ Differentiating w.r.t. m shows that $\dfrac{df}{dm}=0$ when $$3m^2-2m-3=0\\ m=\frac{1\pm\sqrt{10}}3\quad \Rightarrow m^2=\frac{11\pm2\sqrt{10}}9$$ Substituting into the above formula shows that $l^2$ (and hence $l$) is maximum when $$m=\frac{1-\sqrt{10}}3\qquad \blacksquare$$ hence this is the incline (or rather, $\tan\theta$, where $\theta$ is the angle of incline) of the major axis of the ellipse.

3. Semi-major and Semi-minor Axes

The semi-major and semi-minor axes (usually denoted by $a,b$ in the standard ellipse notation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$) are given by $l_{\text{max}},l_{\text{min}}$ respectively.

Hence, semi-major axis, $a$, is

$$a=l_{\text{max}}=\sqrt{\frac7{12}\left(\frac{1+m^2}{3+6m+5m^2}\right)}\biggr|_{m =\frac{1-\sqrt{10}}3} =\sqrt{\frac7{72}(4+\sqrt{10})}$$

and the semi-minor axis, $b$, is

$$b=l_{\text{min}}=\sqrt{\frac7{12}\left(\frac{1+m^2}{3+6m+5m^2}\right)}\biggr|_{m =\frac{1+\sqrt{10}}3} =\sqrt{\frac7{72}(4-\sqrt{10})}\qquad \blacksquare$$

Answer 3

Here's another method suggested by a friend of mine.

Keep the solution of the centre of the ellipse identified earlier as $(-\frac13,\frac12)$ .

Change the centre of the ellipse such to the origin. Equation of translated ellipse is:
$$E':\qquad 3x^2+6xy+5y^2-\frac7{12}=0\qquad \cdots (1)$$ Consider a circle at the same centre as the ellipse (i.e. the origin) with equation $$C': \qquad x^2+y^2=r^2\qquad \qquad \qquad \qquad \cdots (2) $$ At intersection of $E'$ and $C$:

$$(1)-3\times(2):\\ 6xy+2y^2=\frac7{12}-3r^2$$ Substitute in $(1)$: $$\begin{align} 6\sqrt{r^2-y^2}y&=-\left[2y^2+\left((3r^2-\frac7{12}\right)\right]\\ 36(r^2-y^2)y^2&=4y^4+4\left(3r^2-\frac7{12}\right)y^2+\left(3r^2-\frac7{12}\right)^2\\ 40y^4+4\left(-6r^2-\frac7{12}\right)y^2+\left(3r^2-\frac7{12}\right)^2&=0 \end{align}$$ For tangency, $$\begin{align} 4^2\left(6r^2+\frac7{12}\right)^2&=4(40)\left(3r^2-\frac7{12}\right)^2\\ 6r^2+\frac7{12}&=\sqrt{10}\left(3r^2-\frac7{12}\right)\\ 3r^2(2-\sqrt{10})&=\frac7{12}(-1-\sqrt{10})\\ r^2&=\frac7{36}\frac{\sqrt{10}+1}{\sqrt{10}-2}=\frac7{72}(4+\sqrt{10})\\ r&=\sqrt{\frac7{72}(4+\sqrt{10})}\qquad \text{(semi-major axis)}\qquad \blacksquare \end{align}$$

Correspondingly, $$\qquad\qquad r=\sqrt{\frac7{72}(4-\sqrt{10})}\qquad \text{(semi-minor axis)}\qquad \blacksquare$$

The incline of the ellipse can be calculated easily from values of $x,y$ obtained by substituting values of $r$.