Characteristics of the conic $14x^2-4xy+11y^2-44x-58y+71=0$

Question

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?

Answer 1

Let $f(x,y)=14x^2 - 4xy + 11y^2 - 44x - 58y + 71$.

i) The equations to solve for the centre is $f’_x=f’_y=0$, or

$$28x-4y-44=0,\>\>\>\>\>-4x+22y-58=0$$ which yields the center $(2,3)$.

ii) The axes are parallel to the normal vectors at the vertexes, i.e.

$$\frac{f’_y}{f’_x}= \frac{y-3}{x-2}$$

which leads to the respective equations of the major and minor axes

$$2x-y=1, \>\>\>\>\> x+2y=8 $$

and the corresponding major vertexes $\left(2\pm\sqrt{ \frac65} 3\pm 2\sqrt{ \frac65}\right)$ and minor vertexes $\left(2\pm \frac4{\sqrt5}, 3\pm \frac4{\sqrt5}\right) $. Then, the lengths of the axes are $2a=2\sqrt6$ and $2b=4$.

iii). The directrices are parallel to the minor axis $x+2y=8$ and at the distance $\frac{a^2}c= 3\sqrt2$, i.e.

$$\frac{|x+2y-8|}{\sqrt{1^2+2^2}}= 3\sqrt2\implies x+2y= 8\pm 3\sqrt{10}$$

Answer 2

$ 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.

Answer 3

Hints. To show that it's an ellipse. Consider as a quaratic equation in $x$ and show that the condition that $x,\,y$ be real gives that $\alpha\le y\le\beta$ for some $\alpha\ne\beta.$ Similarly show that for some $m\ne n,$ we have that $m\le x\le n $ This shows that the curve is bounded -- only ellipses of the conics satisfy this condition. To rule out the case of the point --the circle with vanishing radius -- show that there are at least two points satisfying the equation -- this is easy to do. Set $x=0$ and see that you will have two distinct values for $y.$

Now, what I'd do is to rotate axes. That is, find a suitable $\phi$ so that making the substitution $$x=X\cos\phi-Y\sin\phi\\y=X\sin\phi+Y\cos\phi$$ gets rid of the term in $xy.$ Afterwards, You will have an equation which may be put into the form $$A(X-a)^2+B(Y-b)^2=1.$$ Then shift the origin to get an equation in recognisable form. From here all problems problems should be solved. So they only want you to rotate axes and change origin!