Show that the conic represented by the equation
$$14x^2 – 4xy + 11y^2 – 44x – 58y + 71=0$$
is an ellipse.
Also find
i). the equation of ellipse referred to the centre as origin
ii). equations of axes and length of axes
iii). directrices.
My Attempt:
Given equation is:
$$14x^2 – 4xy + 11y^2 – 44x – 58y + 71=0$$
Comparing this with $ax^2+2hxy+by^2+2gx+2fy+c=0$ and calculating
$$\Delta = abc+2fgh – af^2 – bg^2 – ch^2$$ gives $\Delta = -9000 \neq 0$.
Also,
$h^2=4$ and $ab=14\times 11$. As $\Delta neq 0$ and $h^2 < ab$ ,the given equation represents an ellipse.
The coordinates of centre can be obtained by solving the equations
$\frac {\partial S}{\partial x} = 0$ and $\frac {\partial S}{\partial y}=0$ where $S=14x^2 – 4xy + 11y^2 – 44x – 58y + 71=0$.
i.e $28x-4y-44=0$ and $-4x+22y-58=0$.
Solving these equations we get:
$x=2$ and $y=3$. How to solve further?
Best Answer
$ 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.