If conic is represented by $21x^2 -6xy +29y^2 +6x-58y-151=0$, then find the centre, length of axes and eccentricity

Question

Using partial derivatives, I found centre of the conic as $(0,1)$ and I think the conic represents an ellipse. But I am not able to find the rest of the answers.

Answer 1

Let $$f(x,y)= 21x^2 -6xy +29y^2 +6x-58y-151 $$ $$f_x’(x,y)= 42x-6y+6$$ $$f_y’(x,y)= 6x+58y-58$$ The center $(0,1)$ is obtained via $f’_x=f’_y=0$ and by matching the normal vectors at vertexes, i.e. $$f’_x:f’_y=(x-0):(y-1)$$ the equations for the major and minor axes are obtained $$3y-x-3=0,\>\>\>\>\>y+3x-1=0$$ Then, substitute them into $f(x,y)= 0$ to get the major vertexes $(\pm\frac9{\sqrt{10}},1\pm\frac3{\sqrt{10}})$ and the minor vertexes $(\mp\sqrt{\frac35},1\pm3\sqrt{\frac35})$, and in turn their respective lengths $2a=6$ and $2b=2\sqrt6$. Thus, the eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}= \frac1{\sqrt3}$.

Answer 2

The fact that there is an $xy$ term means that the axes of the conic are rotated from the coordinate axes. So the first thing I would do is rotate to new $x'y'$ at some angle $\theta$: $$x= x'\cos(\theta)- y'\sin(\theta), \quad y= x'\sin(\theta)+ y'\cos(\theta)$$

\begin{align} 21x^2&= 21(x'^2\cos^2(\theta)- 2x'y'\sin(\theta)\cos(\theta)+ y'^2\sin^2(\theta))\\ -6xy&= -6(x'^2\cos(\theta)\sin(\theta)+ x'y'(\cos^2(\theta)- \sin^2(\theta))- y'^2\sin(\theta)\cos(\theta))\\ 29y^2&= 29(x'^2\sin^2(\theta)+ 2x'y'\sin(\theta)\cos(\theta)+y'^2\cos^2(\theta)\\\\ 21x^2- 6xy+ 29y^2&= (21\cos^2(\theta)- 6\cos(\theta)\sin(\theta)+ 29\sin^2(\theta))x'^2\\&\qquad+ (-6\cos^2(\theta)+ 6\sin^2(\theta)+ 21\sin(\theta)\cos(\theta)+ 58\sin(\theta)\cos(\theta))x'y'\\ &\qquad+ (21\sin^2(\theta)+ 6\sin(\theta)\cos(\theta)+ \cos^2(\theta)y'^2 \end{align}

The point is to eliminate the coefficient of $x'y'$ so we must have $$-6\cos^2(\theta)+ 6\sin^2(\theta)+ 21\sin(\theta)\cos(\theta)+ 58\sin(\theta)\cos(\theta)= 0$$ or $$6\sin^2(\theta)+ 69\sin(\theta)\cos(\theta)- 6\cos^2(\theta)= 0$$

We can think of that as the quadratic equation $6X^2+ 69XY- 6Y^2= 0$. By the quadratic formula $$X= \frac{-69Y\pm\sqrt{4761Y^2+ 144Y^2}}{12}= \frac{-69Y\pm Y\sqrt{4905}}{12}= \frac{1.03}{12}Y.$$ So $\sin(\theta)= \dfrac{1.03}{12}cos(\theta)$ and $\tan(\theta)= \dfrac{1.03}{12}$.

Answer 3

Since Quanto has given an answer using calculus, I'll give an answer using linear algebra alone. First, we may express this conic in terms of matrix multiplication:

$$21x^2 -6xy +29y^2 +6x-58y-151=v^\top M v=0$$ where $v=(x,y,1)^\top$ and

$$M=M^\top=\begin{pmatrix} 21 & -3 & 3 \\ -3 & 29 & -29 \\ 3 & -29 & -151\end{pmatrix}.$$

The presence of nonzero off-diagonal entries in the last row and last column signals that it's not centered, since these lead to the linear terms in the equation. These may be eliminated by adding the second row to the third and then the second column to the third. These are jointly implemented by a similarity transformation with the appropriate upper-triangular matrix $U$:

$$M'=U^\top M U = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix} \begin{pmatrix} 21 & -3 & 3 \\ -3 & 29 & -29 \\ 3 & -29 & -151\end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix} 21 & -3 & 0 \\ -3 & 29 & 0 \\ 0 & 0 & -180\end{pmatrix}.$$

We can now diagonalize $M'$ by hand. By inspection, it has an eigenvalue $-180$. The remaining $2$-by-$2$ block has determinant $600$ and trace $50$, and thus we deduce that the remaining eigenvalues are $30,20$. Computing the eigenvectors by inspection, we deduce that $M''$ is diagonalized as $M'=R \Lambda R^{-1}$ where $\Lambda=\operatorname{diag}(30,20,-180)$ and $$R=\begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}}&0 \\ -\frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} & 0 \\ 0 & 0 & 1\end{pmatrix}.$$ Note that the columns of $R$ are the orthonormal eigenvectors of $M'$ and thus $R^\top=R^{-1}$. Putting these together, we have \begin{align} v^\top M v &=v^\top (U^{-1})^\top M' U^{-1} v\\ &=(U^{-1}v)^\top R \Lambda R^{-1} U^{-1} v\\ &=(R^{-1}U^{-1}v)^\top \Lambda (R^{-1} U^{-1})v\\ &=u^\top \Lambda u \end{align} where $u=R^{-1}U^{-1} v=(UR)^{-1}v$ gives the transformation to a coordinate system in which the ellipse is aligned with the axes. Explicitly, we have $$u=R^{-1}U^{-1}v =\begin{pmatrix} \frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{10}}&0 \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix} x \\ y \\ 1\end{pmatrix} =\begin{pmatrix} \frac{1}{\sqrt{10}}(x-3y+3) \\ \frac{1}{\sqrt{10}}(3x+y-1) \\ 1\end{pmatrix}.$$ We may thus write $u=(x',y',1)^\top$. In terms of the new coordinates, we have $$30x'^2+20y'^2=180\implies\frac{x'^2}{6}+\frac{y'^2}{9}=1.$$ The lengths of the axes are evidently $\sqrt{6}$ and $3$ respectively, and the eccentricity is $\sqrt{1-6/9}=1/\sqrt{3}$. Since the new coordinates are obtained by first translating and then rotating the system, this transformation doesn't change the dimensions and therefore these values also describe the original ellipse. In addition, we may note that the point $(x,y)=(0,1)$ is mapped to $(x',y')=(0,0)$ which we recognize as the center of the ellipse in these coordinates. Hence $(0,1)$ is indeed the center of the original ellipse.

P.S. I can't resist doing a little bit of calculus. Writing $f(x,y)=v^\top M v$ as above, we may write $f_x=2 v^\top M e_1$ and $f_y=2v^\top Me_2$. The condition that $f_x=f_y=0$, then, corresponds to choosing $v=(x,y,1)^\top$ such that it's orthogonal to both $Me_1=(21,-3,3)^\top$ and $Me_2=(-3,29,-29)$. This is readily accomplished by taking the cross product of these last two vectors, obtaining $$(Me_1)\times (Me_2) = (0,600,600).$$ From this we deduce that the vector $(0,1,1)$ will suffice and therefore the center is at $(0,1)$.