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Originally posted by Richard Lee
Let point P(asec(theta), btan(theta)) and Q(asec(Phi), btan(phi))
Therefore, the line passes through P and Q:
[y-btan(theta)]/[x-asec(theta)]=[btan(phi)-btan(theta)]/[asec(phi)-asec(theta)]
After that, sub (ae, 0) and (-ae, 0) into the equation, you will get the answer.
Don't forget the trig formula!
Thanx!Originally posted by Newbie
dude its an ellipse.....
I will have to think about it ...Originally posted by Newbie
ahh ok
anyone wana show me how to do it?![]()
na actually, why don't u check up with the hsc advice line?Originally posted by Fosweb
yeah - thats what our teacher said too. had to look them up. maybe i should learn them anyway? cant hurt.
so i dont know if they'd ask a question like this.
lol: i was thinking about doing that. later.Originally posted by freaking_out
na actually, why don't u check up with the hsc advice line?![]()
he's got one of those curve sketching templates.Originally posted by ND
Fosweb: what did you use to draw that ellipse? Mine always come out looking completely retarded (as with anything else i draw).