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

$ 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 2

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!

Answer 3

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.

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