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(x2y2)+x2+y2a2=0xy(x^2-y^2)+x^2+y^2-a^2=0

xy(x2y2)+x2+y2=a2xy(x^2-y^2)+x^2+y^2=a^2


Let us find asymptotes

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

Substituting gives

x(kx+b)(x2(kx+b)2)+x2+(kx+b)2=a2x(kx+b)(x^2-(kx+b)^2)+x^2+(kx+b)^2=a^2

x3(kx+b)x(kx+b)3+x2+(kx+b)2=a2x^3(kx+b)-x(kx+b)^3+x^2+(kx+b)^2=a^2

kx4+bx3k3x43k2x3b3kx2b2b3x+x2+k2x2+2kx+b2=a2kx^4+bx^3-k^3x^4-3k^2x^3b-3kx^2b^2-b^3x+x^2+k^2x^2+2kx+b^2=a^2

(kk3)x4+(b3k2b)x3+(1+k23kb2)x2+(2k3kb2)x2+b2=a2(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

{kk3=0b(13k2)=0\begin{cases} k-k^3=0\\ b(1-3k^2)=0 \end{cases}

{k(1k2)=0b=0 or 13k2=0\begin{cases} k(1-k^2)=0\\ b=0 \ or \ 1-3k^2=0 \end{cases}

{k=0 or 1k2=0b=0 or 13k2=0\begin{cases} k=0 \ or \ 1-k^2=0\\ b=0 \ or \ 1-3k^2=0 \end{cases}

{k=0 or k=±1b=0 or k=±13\begin{cases} k=0 \ or \ k=\pm1\\ b=0 \ or \ k=\pm\frac{1}{\sqrt{3}} \end{cases}

{k=0 or k=±1b=0\begin{cases} k=0 \ or \ k=\pm1\\ b=0 \end{cases}

y=0y=0 is a horizontal asymptote.

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

2) x=cx=c -vertical asymptote

cy(c2y2)+c2+y2=a2cy(c^2-y^2)+c^2+y^2=a^2

cy3+y2+c3y+c2=a2cy^3+y^2+c^3y+c^2=a^2

c=0c=0

x=0x=0 - vertical asymptote


Using polar coordinates gives

x=ρcosθx=\rho\cos{\theta}

y=ρsinθy=\rho\sin{\theta}

We get

ρcosθρsinθ(ρ2cos2θρ2sin2θ)+ρ2=a2\rho\cos{\theta}\rho\sin{\theta}(\rho^2\cos^2{\theta}-\rho^2\sin^2{\theta})+\rho^2=a^2

ρ2ρ2sinθcosθ(cos2θsin2θ)+ρ2=a2\rho^2\rho^2\sin{\theta}\cos{\theta}(\cos^2{\theta}-\sin^2{\theta})+\rho^2=a^2

ρ412sin2θcos2θ+ρ2=a2\rho^4\frac{1}{2}\sin{2\theta}\cos{2\theta}+\rho^2=a^2

ρ44sin4θ+ρ2=a2\frac{\rho^4}{4}\sin{4\theta}+\rho^2=a^2

sin4θ4ρ4+ρ2=a2\frac{\sin{4\theta}}{4}\rho^4+\rho^2=a^2

If θ=πn\theta=\pi n , then ρ2=a2=>x2+y2=a2\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=±ax=0=>y=\pm a and y=0=>x=±ay=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 x2+y2=a2x^2+y^2=a^2




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