Question

Need help finding asymptotes of curve!

Answer 1

\[\LARGE y^2(x-2a)=x^3-a^3\]

Answer 2

@ganeshie8

Answer 3

@dan815

Answer 4

difference of cubes

Answer 5

eh?

Answer 6

x^3 - a^3 = (x-a)(x^2 +xa + a^2)

Answer 7

yep right

Answer 8

so what difference does it make?

Answer 9

asymptote is a line that approaches a given curve but does not meet it at at any finite distance

Answer 10

yep

Answer 11

take the implicit derivative

Answer 12

Given
y^2(x-2a) = x^3 - a^3

2 y * y ' (x-2a) + y^2 ( 1) = 3x^2

solve for y '

Answer 13

ooo

Answer 14

y' = ( 3x^2 - y^2 ) / (2 y * (x-a))

Answer 15

you have horizontal asymptote when y ' = 0 (numerator = 0) , and vertical asymptote when y ' is undefined (denominator = 0 )

Answer 16

The approach I Follow is:
The variables having nth degree,I replace x with 1 and y with m and solve for phi n(m) and equate it to 0 \[\Large \phi_n (m)= 0\]
this gives em some values of m...lets assume they are distinct.
The value of c is given by :
\[\Large c =\frac{ - \phi_2(m)}{\phi_3'(m)}\]
then y=mx+c
just in case if someones has heard of a method like this....
but im not getting answer by this method

Answer 17

^^im assuming nth degree to be 3 here,so 2 is n-1..actual formula is -phi(n-1)/phi ' (n)

Answer 18

what about y=mx+b asymtotes

Answer 19

how about taking y'' and saying as they approach the asymtotes the change is approaching 0

Answer 20

@perl > the answer is r sin theta=a
i don't think we would get something like that

Answer 21

which is same as y = a hmm

Answer 22

yes

Answer 23

I am actually looking to solve these particular types of questions involving r and theta.
The next question is also similar,
\[\Large r \theta=a\]

Answer 24

but isnt it strange that to have question in rectangular and answer in polar ?

Answer 25

yep that's what strange..the next 5-6 questions have the same pattern!

Answer 26

using ur formula im getting m=1 for the slant asymptote

Answer 27

\[y^2(x-2a)=x^3-a^3 \implies \left(\dfrac{y}{x}\right)^2 \left(1-\dfrac{2a}{x}\right) = 1 - \dfrac{a^3}{x^3}\]

taking the limit \(\large x\to \infty\) :

\[\lim\limits_{x\to\infty} \left[ \left(\dfrac{y}{x}\right)^2 \left(1-\dfrac{2a}{x}\right) = 1 - \dfrac{a^3}{x^3}\right]\]

\[\left(m\right)^2 \left(1-0\right) = 1 -0\]

Answer 28

that gives you \(\large m = \pm 1\)
you still need to find \(c\)

Answer 29

so looke like you will be having 2 slant asymptotes

Answer 30

you already have vertical/horizontal asymptotes from perl's work right ?

Answer 31

i dont think it will be done using this method :/ we don't involve polars

Answer 32

you want to work it in polar coordinates is it ?

Answer 33

to reach the final answer...:P

Answer 34

can you use more words, im having hard time figuring out what exactly are u looking for :O