Question

Find the asymptotes the curve x3-4xy2-3x2+12xy-12y2+8x+2y+4=0 help me please

Answer 1

Answer:

There are three possible asymptotes.

2 y - x = 3,   2 y + x = 3 and x = -3.

Step-by-step explanation:

Asymptotes are the straight lines a curve appears to approach as x reaches points near some value or infinity or -infinity. The slope of the curve becomes almost a constant.

x³ - 4 xy² - 3 x² + 12 xy - 12 y² + 8 x + 2 y + 4 = 0

or,  4(x+3) y² - 2(6x+1) y - (x³-3x²+8x+4) = 0

$y=\frac{(6x+1)+-\sqrt{(6x+1)^2+4(x+3)(x^3-3x^2+8x+4)}}{4(x+3)}\\\\Simplify\ the\ terms\\\\y=\frac{6x+1}{4(x+3)}+-\sqrt{\frac{x^2-6x+35}{4}-\frac{500x+1211}{16(x+3)^2}}\\\\y=\frac{6+\frac{1}{x}}{4+\frac{12}{x}}\\\\+-\frac{x}{2}*\sqrt{1-\frac{6}{x}+\frac{35}{x^2}-\frac{500x+1211}{16x^2(x+3)^2}}\\\ So \ x-> \infty, y=\frac{3}{2}+-\frac{x}{2}$

     

Because the denominator is 0 for x = -3, x = -3 is an asymptote.

As y/x as x -> ∞, is +1/2 or -1/2, we get other two asymptotes.

Thus we get the asymptotes as x = -3, 2y-x=3 and 2y+x = 3.