Asymptotes of a hyperbola

Is this the correct solution something doesnot feel right

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With $b=-21$, I don't think this is very much an ellipse. Perhaps a hyperbola... –  Daryl Oct 3 '12 at 8:37
@Daryl ah yes you are right and I stand corrected, so is the working correct? –  JackyBoi Oct 3 '12 at 8:53
And definitely not an eclipse. :-) –  Hans Lundmark Oct 3 '12 at 9:50
Your write-up is haphazard, but it seems you don't mean what you've written. For instance, "$a^2=a$" and "$b^2=b$" are false statements here, but you wind up with the proper values $a=3.16...$ and $b=1.38...$. On the other hand, "$\sqrt{a/b} = \sqrt{1.38.../3.16...}$" is a true statement, but an irrelevant one: the gradient you seek should be simply "$b/a$" ... which happens to be the value $0.43...$ you computed. So, your final answer is correct, but your stream-of-consciousness style makes it impossible to judge whether you know what you're doing or just guessed your way through. –  Blue Oct 3 '12 at 10:11
By the way ... If your notes make sense to you, that's great. Exactly how your particular thoughts bounce around in your particular brain is no one's concern but your own. However, if you're trying to get people to check your work, or help you find flaws in your reasoning, then you really must make an effort to express those thoughts in a passably-standard way. It's like how twins sometimes develop their own language to communicate with each other; that's terrific --fascinating, really-- but if they ever need driving directions, they're going to want to ask for them in the local dialect. –  Blue Oct 3 '12 at 10:40

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:

$a=\pm\sqrt{10}$, $\ b=\pm\sqrt{40/21}$
$\displaystyle\frac{b}{a}=\pm\sqrt{4/21}\simeq 0.43644$