Question

determine the asymptotes of the curve, x³+3x²y-4y³-x+y+3=0. ​

Answer 1

Answer:

The asymptotes are y-x=0 and 2 y+x=0.

Step-by-step explanation:

A line that is an asymptote of a curve is one where one or both of the x or y coordinates tend to infinity and the distance between the curve and the line approaches zero.

OR, A line that is tangent to a curve at a point at infinity is said to be the asymptote of that curve.

Calculation:

Given curve:

$x^3+3 x^2 y-4 y^3-x+y+3=0$

Step 1: Divide through by $y^3$ :

$\left(\frac{x}{y}\right)^3+3\left(\frac{x}{y}\right)^2-4-\frac{x}{y^3}+\frac{1}{y^2}+\frac{3}{y^2}=0$

When y is very large, the last 3 terms become small

$\Rightarrow\left(\frac{x}{y}\right)^3+3\left(\frac{x}{y}\right)^2-4=0$approximately, and $(\mathrm{x} / \mathrm{y})=1$,

$\therefore \mathrm{y}=\mathrm{x}$ is an asymptote.

Step 2: Similarly, dividing through by $\mathrm{x}^{\wedge} 3$

$\Rightarrow 1+3\left(\frac{y}{x}\right)-4\left(\frac{y}{x}\right)^3=0, \mathrm{y}=\mathrm{x}$ is a solution to this equation

Let $z=y / x$, then $1+3 z-4 z^3=0$, and $z=1$ is a root.

Step 3: Solving we get $(2 z+1)^2$ is the other factor.

$\Rightarrow z=-1 / 2$ is the other root

$\therefore y / x=-1 / 2$ and $y=-x / 2$ are the required asymptote.

∴ The asymptotes are y-x=0 and 2 y+x=0.

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