x y ( x 2 − y 2 ) + x 2 + y 2 − a 2 = 0 xy(x^2-y^2)+x^2+y^2-a^2=0 x y ( x 2 − y 2 ) + x 2 + y 2 − a 2 = 0
x y ( x 2 − y 2 ) + x 2 + y 2 = a 2 xy(x^2-y^2)+x^2+y^2=a^2 x y ( x 2 − y 2 ) + x 2 + y 2 = a 2
Let us find asymptotes
1) y = k x + b y=kx+b y = k x + b - oblique or horizontal asymptote
Substituting gives
x ( k x + b ) ( x 2 − ( k x + b ) 2 ) + x 2 + ( k x + b ) 2 = a 2 x(kx+b)(x^2-(kx+b)^2)+x^2+(kx+b)^2=a^2 x ( k x + b ) ( x 2 − ( k x + b ) 2 ) + x 2 + ( k x + b ) 2 = a 2
x 3 ( k x + b ) − x ( k x + b ) 3 + x 2 + ( k x + b ) 2 = a 2 x^3(kx+b)-x(kx+b)^3+x^2+(kx+b)^2=a^2 x 3 ( k x + b ) − x ( k x + b ) 3 + x 2 + ( k x + b ) 2 = a 2
k x 4 + b x 3 − k 3 x 4 − 3 k 2 x 3 b − 3 k x 2 b 2 − b 3 x + x 2 + k 2 x 2 + 2 k x + 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 x 4 + b x 3 − k 3 x 4 − 3 k 2 x 3 b − 3 k x 2 b 2 − b 3 x + x 2 + k 2 x 2 + 2 k x + b 2 = a 2
( k − k 3 ) x 4 + ( b − 3 k 2 b ) x 3 + ( 1 + k 2 − 3 k b 2 ) x 2 + ( 2 k − 3 k b 2 ) x 2 + 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 ( k − k 3 ) x 4 + ( b − 3 k 2 b ) x 3 + ( 1 + k 2 − 3 k b 2 ) x 2 + ( 2 k − 3 k b 2 ) x 2 + b 2 = a 2
Equalization of coefficients for the two leading terms of the equation gives
{ k − k 3 = 0 b ( 1 − 3 k 2 ) = 0 \begin{cases}
k-k^3=0\\
b(1-3k^2)=0
\end{cases} { k − k 3 = 0 b ( 1 − 3 k 2 ) = 0
{ k ( 1 − k 2 ) = 0 b = 0 o r 1 − 3 k 2 = 0 \begin{cases}
k(1-k^2)=0\\
b=0 \ or \ 1-3k^2=0
\end{cases} { k ( 1 − k 2 ) = 0 b = 0 or 1 − 3 k 2 = 0
{ k = 0 o r 1 − k 2 = 0 b = 0 o r 1 − 3 k 2 = 0 \begin{cases}
k=0 \ or \ 1-k^2=0\\
b=0 \ or \ 1-3k^2=0
\end{cases} { k = 0 or 1 − k 2 = 0 b = 0 or 1 − 3 k 2 = 0
{ k = 0 o r k = ± 1 b = 0 o r k = ± 1 3 \begin{cases}
k=0 \ or \ k=\pm1\\
b=0 \ or \ k=\pm\frac{1}{\sqrt{3}}
\end{cases} { k = 0 or k = ± 1 b = 0 or k = ± 3 1
{ k = 0 o r k = ± 1 b = 0 \begin{cases}
k=0 \ or \ k=\pm1\\
b=0
\end{cases} { k = 0 or k = ± 1 b = 0
y = 0 y=0 y = 0 is a horizontal asymptote.
y = x a n d y = − x y=x\ and\ y=-x y = x an d y = − x are oblique asymptotes
2) x = c x=c x = c -vertical asymptote
c y ( c 2 − y 2 ) + c 2 + y 2 = a 2 cy(c^2-y^2)+c^2+y^2=a^2 cy ( c 2 − y 2 ) + c 2 + y 2 = a 2
c y 3 + y 2 + c 3 y + c 2 = a 2 cy^3+y^2+c^3y+c^2=a^2 c y 3 + y 2 + c 3 y + c 2 = a 2
c = 0 c=0 c = 0
x = 0 x=0 x = 0 - vertical asymptote
Using polar coordinates gives
x = ρ cos θ x=\rho\cos{\theta} x = ρ cos θ
y = ρ sin θ y=\rho\sin{\theta} y = ρ sin θ
We get
ρ cos θ ρ sin θ ( ρ 2 cos 2 θ − ρ 2 sin 2 θ ) + ρ 2 = a 2 \rho\cos{\theta}\rho\sin{\theta}(\rho^2\cos^2{\theta}-\rho^2\sin^2{\theta})+\rho^2=a^2 ρ cos θ ρ sin θ ( ρ 2 cos 2 θ − ρ 2 sin 2 θ ) + ρ 2 = a 2
ρ 2 ρ 2 sin θ cos θ ( cos 2 θ − sin 2 θ ) + ρ 2 = a 2 \rho^2\rho^2\sin{\theta}\cos{\theta}(\cos^2{\theta}-\sin^2{\theta})+\rho^2=a^2 ρ 2 ρ 2 sin θ cos θ ( cos 2 θ − sin 2 θ ) + ρ 2 = a 2
ρ 4 1 2 sin 2 θ cos 2 θ + ρ 2 = a 2 \rho^4\frac{1}{2}\sin{2\theta}\cos{2\theta}+\rho^2=a^2 ρ 4 2 1 sin 2 θ cos 2 θ + ρ 2 = a 2
ρ 4 4 sin 4 θ + ρ 2 = a 2 \frac{\rho^4}{4}\sin{4\theta}+\rho^2=a^2 4 ρ 4 sin 4 θ + ρ 2 = a 2
sin 4 θ 4 ρ 4 + ρ 2 = a 2 \frac{\sin{4\theta}}{4}\rho^4+\rho^2=a^2 4 s i n 4 θ ρ 4 + ρ 2 = a 2
If θ = π n \theta=\pi n θ = πn , then ρ 2 = a 2 = > x 2 + y 2 = a 2 \rho^2=a^2=>x^2+y^2=a^2 ρ 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 = ± a x=0=>y=\pm a x = 0 => y = ± a and y = 0 = > x = ± a y=0=>x=\pm a y = 0 => x = ± 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 x^2+y^2=a^2 x 2 + y 2 = a 2
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