what is asymtote og function y^2 (x−2a)=x^3 −a^3 ?


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Question Number 81195 by jagoll last updated on 10/Feb/20

$w h a t i s a s y m t o t e o g f u n c t i o n$ $y^{2} (x - 2 a) = x^{3} - a^{3} ?$
Commented by mr W last updated on 10/Feb/20
$my method: y^2 (x−2a)=x^3 −a^3 f(x)=y=±(√((x^3 −a^3 )/(x−2a))) say the asymtote for f(x)=(√((x^3 −a^3 )/(x−2a))) is y=px+q f(x)−(px+q)=(√((x^3 −a^3 )/(x−2a)))−px−q let t=x−2a x=t+2a x^3 =t^3 +6at^2 +12a^2 t+8a^3 x^3 −a^3 =t^3 +6at^2 +12a^2 t+7a^3 ((x^3 −a^3 )/(x−2a))=t^2 (1+((6a)/t)+((12a^2 )/t^2 )+((7a^3 )/t^3 )) lim_(x→∞) ((√((x^3 −a^3 )/(x−2a)))−px−q)=0 lim_(t→∞) [t(√(1+((6a)/t)+((12a^2 )/t^2 )+((7a^3 )/t^3 )))−pt−2pa−q]=0 lim_(t→∞) {t[1+((3a)/t)+o((1/t))]−pt−2pa−q}=0 ⇒p=1 ⇒3a−2pa−q=0 ⇒q=a i.e. the asymyote for f(x)=(√((x^3 −a^3 )/(x−2a))) is y=x+a similarly the asymyote for f(x)=−(√((x^3 −a^3 )/(x−2a))) is y=−x−a ⇒the asymtotes for y^2 (x−2a)=x^3 −a^3 are y=x+a and y=−x−a$
$m y m e t h o d :$ $y^{2} (x - 2 a) = x^{3} - a^{3}$ $f (x) = y = \pm \sqrt{\frac{x^{3} - a^{3}}{x - 2 a}}$ $s a y t h e a s y m t o t e f o r f (x) = \sqrt{\frac{x^{3} - a^{3}}{x - 2 a}} i s$ $y = p x + q$ $f (x) - (p x + q) = \sqrt{\frac{x^{3} - a^{3}}{x - 2 a}} - p x - q$ $l e t t = x - 2 a$ $x = t + 2 a$ $x^{3} = t^{3} + 6 {a t}^{2} + 12 a^{2} t + 8 a^{3}$ $x^{3} - a^{3} = t^{3} + 6 {a t}^{2} + 12 a^{2} t + 7 a^{3}$ $\frac{x^{3} - a^{3}}{x - 2 a} = t^{2} (1 + \frac{6 a}{t} + \frac{12 a^{2}}{t^{2}} + \frac{7 a^{3}}{t^{3}})$ $\lim_{x \to \infty} (\sqrt{\frac{x^{3} - a^{3}}{x - 2 a}} - p x - q) = 0$ $\lim_{t \to \infty} [t \sqrt{1 + \frac{6 a}{t} + \frac{12 a^{2}}{t^{2}} + \frac{7 a^{3}}{t^{3}}} - p t - 2 p a - q] = 0$ $\lim_{t \to \infty} {t [1 + \frac{3 a}{t} + o (\frac{1}{t})] - p t - 2 p a - q} = 0$ $\Rightarrow p = 1$ $\Rightarrow 3 a - 2 p a - q = 0 \Rightarrow q = a$ $i . e . t h e a s y m y o t e f o r f (x) = \sqrt{\frac{x^{3} - a^{3}}{x - 2 a}}$ $i s y = x + a$ $s i m i l a r l y t h e a s y m y o t e f o r f (x) = - \sqrt{\frac{x^{3} - a^{3}}{x - 2 a}}$ $i s y = - x - a$ $\Rightarrow t h e a s y m t o t e s f o r y^{2} (x - 2 a) = x^{3} - a^{3} a r e$ $y = x + a a n d y = - x - a$
Commented by mr W last updated on 10/Feb/20

Commented by jagoll last updated on 10/Feb/20

$s i r v e r t i c a l a s y m t o t e i s x = 2 a$ $i t c o r r e c t ?$
Commented by jagoll last updated on 10/Feb/20

$t h a n k y o u v e r y m u c h s i r$
Commented by ajfour last updated on 10/Feb/20

$y = 0 a t x = a, x = 2 a i s v e r t i c a l$ $a s y m t o t e . I n y o u r d i a g r a m, S i r,$ $i t s e e m s a = 2 a! ?$
Commented by mr W last updated on 10/Feb/20

Commented by mr W last updated on 10/Feb/20

$t h i s i s a n e x a m p l e w i t h a = 5 .$ $v e r t i c a l a s y m t o t e i s x = 2 a = 10 .$ $x = 5 a t y = 0 .$
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