Answer to Question #153129 in Calculus for Neha

Question

Find the asymptotes - x3 - 2x2y - 2x2 + 4xy - 4y2 + 4x + y + 3 = 0

Answer 1

The given equation of the curve is $-x^3-2x^2y-2x^2+4xy-4y^2+4x+y+3=0$

We shall first find the vertical asymptotes of the if any :

The highest power of the term of $y$ is $y^2.$ Its coefficient is $-4$ (constant).

So the curve has no vertical asymptotes.

Next we shall find horizontal asymptote:

Let $y=mx+c$ be the horizontal asymptote.

Then $m$ and $c$ can be find in the following way.

We arrange the equation of the curve as

$(-x^3-2x^2y)+(-2x^2+4xy-4y^2)+(4x+y)+3=0$

$\implies x^3[-1-2.\frac{y}{x}]+x^2[-2+4.\frac{y}{x}-4(\frac{y}{x})^2]+(4x+y)+3=0$

Here , $\phi_3(\frac{y}{x})=-1-2(\frac{y}{x})$

$\phi_2(\frac{y}{x})=-2+4(\frac{y}{x})-4(\frac{y}{x})^2$

$\phi _1(\frac{y}{x})=4x+y$

$\phi_0(\frac{y}{x})=3$

Now consider the equation, $\phi_3(m)=0$

$\implies-1-2m=0$

$\implies m=-\frac {1}{2}$

Again we know that , $c=-\frac{\phi_2(m)}{\phi_3'(m)}$ [ if $\phi_3'(m)\neq0$ ]

Here $\phi_3'(m)=-2$ and $\phi_2(m)=-2+4m-4m^2$

Now corresponding to $m=-\frac{1}{2}$ , $c=-\frac{\phi_2(-\frac{1}{2})}{\phi_3'(-\frac{1}{2})}$

$\therefore \phi_2(-\frac{1}{2})=-5$ and $\phi_3'(-\frac{1}{2})=-2$ and $c=-\frac{5}{2}$

Hence $y=(-\frac{1}{2}).x+(-\frac{5}{2})$ is an asymptote.

Therefore $x+2y+5=0$ is the only horizontal asymptote of the given curve.