Find The asymptotes of the curve (a^2/x^2)+(b^2/y^2)=1

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Find The asymptotes of the curve

\frac{a^2}{x^2}+\frac{b^2}{y^2}=1

solution

\frac{a^2}{x^2}+\frac{b^2}{y^2}=1 \Rightarrow \,b^2x^2+a^2y^2=x^2y^2

equating the coefficient of highest powers of x to zero ,

we\,get\, \;b^2-y^2=0\Rightarrow\;y=\pm b

equating the coefficient of highest powers of y to zero ,

we\,get\, \;a^2-x^2=0\Rightarrow\;x=\pm a

\therefore x=\pm \;a \; and \;y=\pm \;b \; are\; the \;required\;asymptotes

r=\frac{3\theta}{\theta+1}

Solution – the equation of the curve is $r=\frac{3\theta}{\theta+1}=f(\theta)\, ,say$

\lim_{\theta\rightarrow \infty}f(\theta)=\lim_{\theta\rightarrow \infty}\frac{3\theta}{\theta+1 }=\lim_{\theta\rightarrow \infty}\frac{3}{1+\frac{1}{\theta} }=3

r=3 circular asymptote

Find the asymptotes parallel to the axes of the curve

x^2 y^2 - x^2 - y^2 - x - y + 1 = 0

Given curve is

x^2 y^2 - x^2 - y^2 - x - y + 1 = 0

Given curve is of degree 4 so we cannot have more than 4 asymptotes

Asymptotes parallel to x axis :

Equating the coefficient of highest power of x $(i.e,\; of\; x^2 \;)$ equals to zero,

We get

y^2-1=0 \implies y^2=1 \implies y =\pm1 \\ \implies y-1=0 \; and \; y+1=0

Asymptotes parallel to y axis :

Equating the coefficient of highest power of y $(i.e,\; of\; y^2 \;)$ equals to zero,

We get

x^2-1=0 \implies (x-1)(x+1)=0 \\ \implies x-1=0 \; and \; x+1=0

The asymptotes of the given curve are

{\color{Blue} y-1=0,\;y+1=0,\; x-1=0 \; and \; x+1=0}

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how to find asymptotes of a curve

asymptotes of a curve