Is this the correct solution something doesnot feel right
Asymptotes of a hyperbola
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$\begingroup$
geometry
asymptotics
conic-sections
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0@Blue oh sorry I just found this method of taking a screenshot of my Galaxynote 10.1 so was excited to share, but totally agree with you that I indeed need to improve my presentation skills, my tots do bounce in my brain. Will heed your advice, tks. – 2012-10-03
1 Answers
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Just like Blue said, you may better care how to write your thoughts on the paper.
Students like to use the $=\,$ sign for more kinds of things, while in math this has a rather strict meaning. So, instead rather use arrow or colon.. The other thing you messed up is $a$ alias $\sqrt{a^2}$ and $\sqrt{a}$. This latter is not needed now, fortunately.
So, once $a^2$ is found/defined to be $10$, it will stay $10$. Your last 3 lines with more exactness should go like this, for example:
(Gradients of the) Asymptotes
$a=\pm\sqrt{10}$, $\ b=\pm\sqrt{40/21}$
$\displaystyle\frac{b}{a}=\pm\sqrt{4/21}\simeq 0.43644$