Question

A conic section is given by the equation 57x^2+14√3 xy+43y^2-576=0, identity and sketch the conics.

Answer 1

If we have an equation $Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0,$ we should convert it into a canonical equation $A\tilde{x_1}^2+C_1\tilde{y}^2+F_1=0.$

Let $S=A+C, \;\;\; \delta = \begin{vmatrix} A & B \\ B & C \end{vmatrix}, \;\;\; \Delta = \begin{vmatrix} A &B & D \\ B &C & E \\ D&E&F \end{vmatrix}.$ In our case $A=57, \;\; B=7\sqrt{3}, \;\; C=43, \;\; F=-576.$ Therefore, $S=100, \;\; \delta = 2304, \;\; \Delta = -1327104.$

Next, we solve the system

$\begin{cases} A_1+C_1 = S, \\ A_1C_1=\delta, \\ A_1C_1F_1 = \Delta. \end{cases} \\ A_1 = 36, C_1= 64, F_1 = -576 \;\;\; \text{or} \;\;\; A_1 = 64, C_1= 36, F_1 = -576.$

So the canonical equation will have form

$36\tilde{x}^2+64\tilde{y}^2=576, \;\; \dfrac{(6\tilde{x})^2}{24^2} + \dfrac{(8\tilde{y})^2}{24^2} = 1, \;\; \dfrac{\tilde{x}^2}{4^2} + \dfrac{\tilde{y}^2}{3^2} = 1.$ Another version of equation is

$\dfrac{\tilde{x}^2}{3^2} + \dfrac{\tilde{y}^2}{4^2} = 1.$

It is an ellipse with semiaxes 4 and 3.

The center can be calculated from system

$\begin{cases} Ax_0+By_0+D=0,\\ Bx_0+Cy_0+E=0. \end{cases}\;\;\; \begin{cases} 57x_0+7\sqrt{3}y_0+0=0,\\ 7\sqrt{3}x_0+43y_0+0=0. \end{cases} \;\;\;x_0 = 0, y_0 = 0.$

The angle of shift of the axes is

$\tan \alpha = \dfrac{A_1-A}{B} = - \sqrt{3}, \; \alpha = -60^\circ.$