Answer to Question #125318 in Analytic Geometry for gyamfi francis

Question #125318

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


1
Expert's answer
2020-07-06T15:34:27-0400

If we have an equation Ax2+2Bxy+Cy2+2Dx+2Ey+F=0,Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0, we should convert it into a canonical equation Ax1~2+C1y~2+F1=0.A\tilde{x_1}^2+C_1\tilde{y}^2+F_1=0.

Let S=A+C,      δ=ABBC,      Δ=ABDBCEDEF.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=73,    C=43,    F=576.A=57, \;\; B=7\sqrt{3}, \;\; C=43, \;\; F=-576. Therefore, S=100,    δ=2304,    Δ=1327104.S=100, \;\; \delta = 2304, \;\; \Delta = -1327104.

Next, we solve the system

{A1+C1=S,A1C1=δ,A1C1F1=Δ.A1=36,C1=64,F1=576      or      A1=64,C1=36,F1=576.\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

36x~2+64y~2=576,    (6x~)2242+(8y~)2242=1,    x~242+y~232=1.36\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

x~232+y~242=1.\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

{Ax0+By0+D=0,Bx0+Cy0+E=0.      {57x0+73y0+0=0,73x0+43y0+0=0.      x0=0,y0=0.\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

tanα=A1AB=3,  α=60.\tan \alpha = \dfrac{A_1-A}{B} = - \sqrt{3}, \; \alpha = -60^\circ.






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