Asymptotes of the Hyperbola

[@zakusa9891]

i like how you derive stuff. i see other youtubers just giving formulas and plugging and chugging

[@celebrity_rooster7488]

Exactly. Just write that shit on paper.

[@deenessop5911]

Hey Eddie. About that proof for the equation of the asymptote, I was just wondering that isn't the multiplication rule for limits: that the lim x-> infinity [f(x)] multiplied by the lim x -> infinity [g(x)] NOT the same as the limit of x-> infinity [f(x).g(x)] I've been taught that multiplication rule for limits only holds true for when x-> a constant value. Is that a correct proof then? Crown Academy > DrDu. Gvng gvng. Eskeetit mums Hehexd

[@jeejee3091]

no what Eddie's multiplication rule is the right one

[@joshgun5311]

The angle between the hyperbola asymptotes is 60 degrees, find the eccentricity, how do we do this question?

[@LimeriuxasLT]

hello nice video but sorry not following how u got 1-(a^2/x^2) at 3:48.. cause we divided.... divided what??

[@TheDanubeDepleter]

He divided the entire radicand by x^2/a^2

[@NadeemShaikh-ot8yl]

This is a very nice trick!! Thanks a lot.

[@warrenw6888]

Isn’t using linear equation the same thing since it’s a straight line?

[@christiancarter255]

Yes.

[@guitarttimman]

That's good reasoning.

[@kevinding1204]

ahhhh I would love to like this video so much but the total amount of likes right now is at 314 and I don't want to ruin it :((((((((((

[@Sobioytccc]

Wonderful explaination 👍

[@zizo-ve8ib]

May I ask where the rest of this part is because I can't find the video where you do the rest of the derivation

[@evajaymusic1322]

I like it

[@ugursoydan8187]

it was very helpful for me. thank you

[@aashishrimal579]

Love from nepal

[@cavelinguam6444]

Reallllly helpful

[@aryanbansal624]

could you please please arrange the videos it is really hard to watch all hyperbola asymptote videos in order please reply dont ignore it is a humble request

[@andeslam7370]

would like to know how to infer -bx/a after knowing bx/a is one of the asymptotes

[@Zosso-1618]

Not sure how helpful this is a year after, but I'll give it a shot. When you square root something, there is both a positive and negative variant. The same is true for here because when he took the square root of both sides, there is both a positive and negative version. Like Mr. Woo said, you can just take bx/a and slap a negative onto it to get the other asymptote. I hope that helps!