The "complete the square" method allows you to center the conic, i.e. make the linear terms vanish (by a translation, $ax^2+bx+c\to a'u^2+c'$).
To deal with "obliqueness", you need to let the cross term $xy$ vanish, by means of a rotation.
Let $x=cu-sv,y=su+cv$, which expresses a rotation around the origin ($c,s$ denote the cosine and sine of the angle), and let the centered equation be in the form
$$Ax^2+2Bxy+Cy^2=1.$$
Then substituting,
$$(Ac^2+2Bcs+Cs^2)u^2+(-2Acs+B(c^2-s^2)+2Ccs)uv+(As^2-2Bcs+Cc^2)v^2=1.$$
By a suitable choice of the angle, you can achieve
$$-2Acs+B(c^2-s^2)+2Ccs=0$$
and the equation reduces to
$$A'u^2+C'v^2=1.$$
Depending on the signs of $A',C'$, you get an ellipse or an hyperbola. By a further rescaling of the variables, you can obtain the "canonical" forms (circle and equilateral hyperbola)
$$p^2\pm q^2=1.$$
To find the suitable angle, use the "double angle" formulas and rewrite
$$(C-A)\sin(2\theta)+B\cos(2\theta)=0,$$ which is easy to solve.
$ 14x^2 -4xy +11y^2 -44x -58y +71 = 0$
Matrix form of this equation is
$ \vec{x}^{t} A \vec{x} +K \vec{x} + 71 = 0 \ \ (1) $
where
$ A =\left[\begin{matrix}14 & -2 \\- 2 & 11 \end{matrix}\right] $
end
$ K =\left[ \begin{matrix} - 44& -58 \end{matrix} \right].$
The characteristic equation of $ A $ is
$ \det(A - \lambda I) = \det \left[\begin{matrix}14-\lambda & -2 \\- 2 & 11-\lambda \end{matrix}\right] = (14 -\lambda)(11-\lambda) - 4 = 0 $
$ \lambda^2 -25\lambda +150 = 0$
so eingenvalues of $ A $ are $ \lambda_{1}= 10, \ \ \lambda_{2}= 15$.
We'll find orthonormal bases for the eigenspaces,
$ \lambda_{1}= 10 $
$\left[\begin{matrix}14-10 & -2 \\- 2 & 11-10 \end{matrix}\right] \left[\begin{matrix}a \\ b \end{matrix}\right] = \left[\begin{matrix}0 \\ 0 \end{matrix}\right]$
$\left[\begin{matrix}4 & -2 \\- 2 & 1 \end{matrix}\right] \left[\begin{matrix}a \\ b \end{matrix}\right] = \left[\begin{matrix}0 \\ 0 \end{matrix}\right]$
$ \begin{cases} 4a -2b = 0 \\ -2a +b = 0 \end{cases}$
$ \vec{v}_{1} = \left[ \begin{matrix} a\\ 2a \end{matrix}\right] = a\left[ \begin{matrix} 1\\ 2 \end{matrix}\right], \ \ a\in R.$
$ \lambda_{2}= 15: $
$\left[\begin{matrix}14-15 & -2 \\- 2 & 11-15 \end{matrix}\right] \left[\begin{matrix}c \\ d \end{matrix}\right] = \left[\begin{matrix}0 \\ 0 \end{matrix}\right]$
$\left[\begin{matrix}c \\ d \end{matrix}\right] = \left[\begin{matrix}0 \\ 0 \end{matrix}\right]$
$ \begin{cases} -c -2d = 0 \\ -2c -4d = 0 \end{cases}$
$ \vec{v}_{2} = \left[ \begin{matrix} -2d\\ d \end{matrix}\right] = d\left[ \begin{matrix} -2\\ 1 \end{matrix}\right], \ \ d\in R.$
Accept $a = d = 1. $
Thus,
$ P =\left[\begin{matrix}\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{matrix}\right]$
orthogonally diagonalizes $ \vec{x}^{t}A \vec{x}$
Substituting $ \vec{x} = P\vec{x'}$ into $(1)$ gives
$(P\vec{x'})^{t}\cdot A \cdot (P\vec{x'}) + K(P\vec{x'}) +71 = 0 $
$ (\vec{x'}^{t})(P^{t}A P)\vec{x'} = K\cdot P \vec{x'} + 71 = 0 \ \ (2) $
Since
$ P^{t}A P = \left[\begin{matrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\-\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{matrix} \right]\cdot \left[\begin{matrix}14 & -2 \\- 2 & 11 \end{matrix}\right] \cdot \left[\begin{matrix}\frac{1}{\sqrt{5}} &-\frac{2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{matrix}\right] = \left[\begin{matrix} 10 & 0 \\ 0 & 15 \end{matrix}\right] $
and
$K\cdot P = \left[ \begin{matrix} - 44& -58 \end{matrix} \right]\cdot \left[\begin{matrix}\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{matrix}\right] = \left[ \begin{matrix} -\frac{160}{\sqrt{5}}& \frac{30}{\sqrt{5}} \end{matrix} \right]$
and
$ (2) $ can be written as
$ 10x'^2 + 15y'^2 -\frac{160}{\sqrt{5}}x' + \frac{30}{\sqrt{5}}y' + 71 = 0 $
To bring the conic into standard position the $ x', \ \ y'$ -axes must be translated
$ 10\left( x'^{2}-\frac{16}{\sqrt{5}}x'\right) + 15\left(y'{^2}+ \frac{2}{\sqrt{5}}y'\right) + 71 = 0 $
Completing the squares yields
$ 10\left(x'^2-2\cdot\frac{8}{\sqrt{5}}x' + \frac{64}{5}\right) - \frac{640}{5} + 15\left(y'^2 + 2\cdot \frac{1}{\sqrt{5}}y' + \frac{1}{5}\right) - \frac{15}{5} + 71 = 0 $
$10 \left(x'-\frac{8}{\sqrt{5}}\right)^2 + 15\left(y' + \frac{1}{\sqrt{5}}\right)^2 -60 = 0 \ \ (3)$
If we translate the coordinate axes by means of translation equations
$ x^{"} = x^{'} - \frac{8}{\sqrt{5}}, \ \ y^{"} = y^{'} + \frac{1}{\sqrt{5}} $
then (3) becomes
$ 10 x''^2 + 15 ''^2 = 60 $
or
$ \frac{x''^2}{6} + \frac{y''^2}{4} = 1, $
which is equation of ellipse.
J want you draw of this ellipse with the directional vectors $ \vec{v_{1}}, \vec{v_{2}}$ and translations.
Please find equations of directrices.
Best Answer
The formula derived in this answer is $$ e^2=\frac{2\sqrt{(A{-}C)^2+B^2}}{\sigma(A{+}C)+\sqrt{(A{-}C)^2+B^2}} $$ Setting $A=39,B=-96,C=11,D=14,E=2,F=-34$ gives $$ \begin{align} e^2&=\frac{2\cdot100}{-50+100}=4\\ e&=2 \end{align} $$ since $\sigma=\operatorname{sign}\left(F\!\left(B^2-4AC\right)+\left(AE^2{-}BDE{+}CD^2\right)\right)=\operatorname{sign}\left(-250000\right)=-1$