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

Question

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

Answer 1

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

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

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

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

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

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

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

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

$y=0$ 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 $x^2+y^2=a^2$