Question

Find all the asymptotes of
(C+y)2(x+2y+2)=x+9y-2

Answer 1

Find all the asymptotes of

$(x+y)^{2} (x+2y+2)=x+9y-2$

Answer:

The asymptotes are $y=-x$ and $y=-x/2$

Step-by-step explanation:

Given: $(x+y)^{2} (x+2y+2)=x+9y-2$

To find: all the  asymptotes of curve.

To begin, rearrange the equation.

$(x+y)^{2} (x+2y+2)=x+9y-2$

We wish to know how the function behaves when x approaches ∞ and -∞.

Dividing equation by $x^{3}$ we get,

$\frac{(x+y)^{2} (x+2y+2)-x+9y-2}{x^{3}} =0$

$(\frac{x+y}{x} )^{2}(\frac{x+2y+2}{x})-\frac{x}{x^{3}} -\frac{9y}{x^{3}}+\frac{2}{x^{3}}=0$

$(1+\frac{y}{x})^{2}(\frac{x+2y+2}{x}) -\frac{x}{x^{3}} -\frac{9y}{x^{3}}+\frac{2}{x^{3}}=0$

$(1+\frac{y}{x})^{2}(1+\frac{2y}{x}+\frac{2}{x})-\frac{1}{x^{2}} -\frac{9y}{x^{3}}+\frac{2}{x^{3}}=0$

Upon letting x→∞ or x→-∞ equation became.

$(1+\frac{y}{x})^{2}(1+\frac{2y}{x} )=0$

Let $w=\frac{y}{x}$ so, $(1+w)^{2}(1+2w)=0$

Solving for w, we get $w=-1$, or $w=\frac{-1}{2}$

Hence,  asymptotes are $y=-x$ and $y=-x/2$

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