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 "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}\n A & B \\\\\n B & C\n\\end{vmatrix}, \\;\\;\\; \\Delta = \\begin{vmatrix}\n A &B & D \\\\\n B &C & E \\\\\nD&E&F\n\\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}\nA_1+C_1 = S, \\\\\nA_1C_1=\\delta, \\\\\nA_1C_1F_1 = \\Delta.\n\\end{cases}\n\\\\\nA_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}\nAx_0+By_0+D=0,\\\\\nBx_0+Cy_0+E=0.\n\\end{cases}\\;\\;\\;\n\\begin{cases}\n57x_0+7\\sqrt{3}y_0+0=0,\\\\\n7\\sqrt{3}x_0+43y_0+0=0.\n\\end{cases}\n\\;\\;\\;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."






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