Question Number 81195 by jagoll last updated on 10/Feb/20
$${what}\:{is}\:{asymtote}\:{og}\:{function} \\ $$$${y}^{\mathrm{2}} \left({x}−\mathrm{2}{a}\right)={x}^{\mathrm{3}} −{a}^{\mathrm{3}} \:? \\ $$
Commented by mr W last updated on 10/Feb/20
![my method: y^2 (x−2a)=x^3 −a^3 f(x)=y=±(√((x^3 −a^3 )/(x−2a))) say the asymtote for f(x)=(√((x^3 −a^3 )/(x−2a))) is y=px+q f(x)−(px+q)=(√((x^3 −a^3 )/(x−2a)))−px−q let t=x−2a x=t+2a x^3 =t^3 +6at^2 +12a^2 t+8a^3 x^3 −a^3 =t^3 +6at^2 +12a^2 t+7a^3 ((x^3 −a^3 )/(x−2a))=t^2 (1+((6a)/t)+((12a^2 )/t^2 )+((7a^3 )/t^3 )) lim_(x→∞) ((√((x^3 −a^3 )/(x−2a)))−px−q)=0 lim_(t→∞) [t(√(1+((6a)/t)+((12a^2 )/t^2 )+((7a^3 )/t^3 )))−pt−2pa−q]=0 lim_(t→∞) {t[1+((3a)/t)+o((1/t))]−pt−2pa−q}=0 ⇒p=1 ⇒3a−2pa−q=0 ⇒q=a i.e. the asymyote for f(x)=(√((x^3 −a^3 )/(x−2a))) is y=x+a similarly the asymyote for f(x)=−(√((x^3 −a^3 )/(x−2a))) is y=−x−a ⇒the asymtotes for y^2 (x−2a)=x^3 −a^3 are y=x+a and y=−x−a](https://www.tinkutara.com/question/Q81211.png)
$${my}\:{method}: \\ $$$${y}^{\mathrm{2}} \left({x}−\mathrm{2}{a}\right)={x}^{\mathrm{3}} −{a}^{\mathrm{3}} \\ $$$${f}\left({x}\right)={y}=\pm\sqrt{\frac{{x}^{\mathrm{3}} −{a}^{\mathrm{3}} }{{x}−\mathrm{2}{a}}} \\ $$$${say}\:{the}\:{asymtote}\:{for}\:{f}\left({x}\right)=\sqrt{\frac{{x}^{\mathrm{3}} −{a}^{\mathrm{3}} }{{x}−\mathrm{2}{a}}}\:{is} \\ $$$${y}={px}+{q} \\ $$$${f}\left({x}\right)−\left({px}+{q}\right)=\sqrt{\frac{{x}^{\mathrm{3}} −{a}^{\mathrm{3}} }{{x}−\mathrm{2}{a}}}−{px}−{q} \\ $$$${let}\:{t}={x}−\mathrm{2}{a} \\ $$$${x}={t}+\mathrm{2}{a} \\ $$$${x}^{\mathrm{3}} ={t}^{\mathrm{3}} +\mathrm{6}{at}^{\mathrm{2}} +\mathrm{12}{a}^{\mathrm{2}} {t}+\mathrm{8}{a}^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} −{a}^{\mathrm{3}} ={t}^{\mathrm{3}} +\mathrm{6}{at}^{\mathrm{2}} +\mathrm{12}{a}^{\mathrm{2}} {t}+\mathrm{7}{a}^{\mathrm{3}} \\ $$$$\frac{{x}^{\mathrm{3}} −{a}^{\mathrm{3}} }{{x}−\mathrm{2}{a}}={t}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{6}{a}}{{t}}+\frac{\mathrm{12}{a}^{\mathrm{2}} }{{t}^{\mathrm{2}} }+\frac{\mathrm{7}{a}^{\mathrm{3}} }{{t}^{\mathrm{3}} }\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{\frac{{x}^{\mathrm{3}} −{a}^{\mathrm{3}} }{{x}−\mathrm{2}{a}}}−{px}−{q}\right)=\mathrm{0} \\ $$$$\underset{{t}\rightarrow\infty} {\mathrm{lim}}\left[{t}\sqrt{\mathrm{1}+\frac{\mathrm{6}{a}}{{t}}+\frac{\mathrm{12}{a}^{\mathrm{2}} }{{t}^{\mathrm{2}} }+\frac{\mathrm{7}{a}^{\mathrm{3}} }{{t}^{\mathrm{3}} }}−{pt}−\mathrm{2}{pa}−{q}\right]=\mathrm{0} \\ $$$$\underset{{t}\rightarrow\infty} {\mathrm{lim}}\left\{{t}\left[\mathrm{1}+\frac{\mathrm{3}{a}}{{t}}+{o}\left(\frac{\mathrm{1}}{{t}}\right)\right]−{pt}−\mathrm{2}{pa}−{q}\right\}=\mathrm{0} \\ $$$$\Rightarrow{p}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{a}−\mathrm{2}{pa}−{q}=\mathrm{0}\:\Rightarrow{q}={a} \\ $$$${i}.{e}.\:{the}\:{asymyote}\:{for}\:{f}\left({x}\right)=\sqrt{\frac{{x}^{\mathrm{3}} −{a}^{\mathrm{3}} }{{x}−\mathrm{2}{a}}} \\ $$$${is}\:{y}={x}+{a} \\ $$$${similarly}\:\:{the}\:{asymyote}\:{for}\:{f}\left({x}\right)=−\sqrt{\frac{{x}^{\mathrm{3}} −{a}^{\mathrm{3}} }{{x}−\mathrm{2}{a}}} \\ $$$${is}\:{y}=−{x}−{a} \\ $$$$ \\ $$$$\Rightarrow{the}\:{asymtotes}\:{for}\:{y}^{\mathrm{2}} \left({x}−\mathrm{2}{a}\right)={x}^{\mathrm{3}} −{a}^{\mathrm{3}} \:{are}\: \\ $$$${y}={x}+{a}\:{and}\:{y}=−{x}−{a} \\ $$
Commented by mr W last updated on 10/Feb/20
Commented by jagoll last updated on 10/Feb/20
$${sir}\:{vertical}\:{asymtote}\:{is}\:{x}=\:\mathrm{2}{a} \\ $$$${it}\:{correct}? \\ $$
Commented by jagoll last updated on 10/Feb/20
$${thank}\:{you}\:{very}\:{much}\:{sir} \\ $$
Commented by ajfour last updated on 10/Feb/20
$${y}=\mathrm{0}\:{at}\:{x}={a}\:,\:{x}=\mathrm{2}{a}\:{is}\:{vertical} \\ $$$${asymtote}.\:{In}\:{your}\:{diagram},\:{Sir}, \\ $$$${it}\:{seems}\:{a}=\mathrm{2}{a}\:!? \\ $$
Commented by mr W last updated on 10/Feb/20
Commented by mr W last updated on 10/Feb/20
$${this}\:{is}\:{an}\:{example}\:{with}\:{a}=\mathrm{5}. \\ $$$${vertical}\:{asymtote}\:{is}\:{x}=\mathrm{2}{a}=\mathrm{10}. \\ $$$${x}=\mathrm{5}\:{at}\:{y}=\mathrm{0}. \\ $$