Answer to Question #93505 in Differential Geometry | Topology for Vanshika

Question #93505
Find the asymtotes of the curve xu(x^2-y^2) +x^2+y^2-a^2.show that the eight points of intersection of the curve with its asymtotes lie on a circle whose centre is at the origin
1
Expert's answer
2019-08-30T08:35:30-0400

"xy(x^2-y^2)+x^2+y^2-a^2=0"

"xy(x^2-y^2)+x^2+y^2=a^2"


Let us find asymptotes

1) "y=kx+b" - oblique or horizontal asymptote

Substituting gives

"x(kx+b)(x^2-(kx+b)^2)+x^2+(kx+b)^2=a^2"

"x^3(kx+b)-x(kx+b)^3+x^2+(kx+b)^2=a^2"

"kx^4+bx^3-k^3x^4-3k^2x^3b-3kx^2b^2-b^3x+x^2+k^2x^2+2kx+b^2=a^2"

"(k-k^3)x^4+(b-3k^2b)x^3+(1+k^2-3kb^2)x^2+(2k-3kb^2)x^2+b^2=a^2"

Equalization of coefficients for the two leading terms of the equation gives

"\\begin{cases}\nk-k^3=0\\\\\nb(1-3k^2)=0\n\\end{cases}"

"\\begin{cases}\nk(1-k^2)=0\\\\\nb=0 \\ or \\ 1-3k^2=0\n\\end{cases}"

"\\begin{cases}\nk=0 \\ or \\ 1-k^2=0\\\\\nb=0 \\ or \\ 1-3k^2=0\n\\end{cases}"

"\\begin{cases}\nk=0 \\ or \\ k=\\pm1\\\\\nb=0 \\ or \\ k=\\pm\\frac{1}{\\sqrt{3}}\n\\end{cases}"

"\\begin{cases}\nk=0 \\ or \\ k=\\pm1\\\\\nb=0\n\\end{cases}"

"y=0" is a horizontal asymptote.

"y=x\\ and\\ y=-x" are oblique asymptotes

2) "x=c" -vertical asymptote

"cy(c^2-y^2)+c^2+y^2=a^2"

"cy^3+y^2+c^3y+c^2=a^2"

"c=0"

"x=0" - vertical asymptote


Using polar coordinates gives

"x=\\rho\\cos{\\theta}"

"y=\\rho\\sin{\\theta}"

We get

"\\rho\\cos{\\theta}\\rho\\sin{\\theta}(\\rho^2\\cos^2{\\theta}-\\rho^2\\sin^2{\\theta})+\\rho^2=a^2"

"\\rho^2\\rho^2\\sin{\\theta}\\cos{\\theta}(\\cos^2{\\theta}-\\sin^2{\\theta})+\\rho^2=a^2"

"\\rho^4\\frac{1}{2}\\sin{2\\theta}\\cos{2\\theta}+\\rho^2=a^2"

"\\frac{\\rho^4}{4}\\sin{4\\theta}+\\rho^2=a^2"

"\\frac{\\sin{4\\theta}}{4}\\rho^4+\\rho^2=a^2"

If "\\theta=\\pi n" , then "\\rho^2=a^2=>x^2+y^2=a^2"

Consequently, points (a, 0), (0, a), (-a, 0), (0, -a) belong to the curve. These points lie on asymptotes.

We could check it sustituting "x=0=>y=\\pm a" and "y=0=>x=\\pm a"

If we build the curve using information we have found and substituting points, we get such curve (for example, a=2):



Let us build asymptotes and circle "x^2+y^2=a^2"




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