|a|e - |a| = |a|(e-1) = 1Can someone please write a solution for 16b
Therefore |a| = 1
a= 1 or a = -1
Oh and also you can do the other vertex:
|a|e + |a| = |a|(1+e) = 1
Therefore |a| = 1/3
a = 1/3 or a = -1/3
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|a|e - |a| = |a|(e-1) = 1Can someone please write a solution for 16b
Yeah, wanna double major in math and physics.Oh wow, I guess I'll retract my previous statement then.
Good on you either way - do you plan on pursuing mathematics in university?
I don't think so- what about vertices on y axis? You have to deal with a quadratic for those. Also you considered only one vertex on the x axis|a|e - |a| = |a|(e-1) = 1
Therefore |a| = 1
a= 1 or a = -1
Am so glad to hear thatYeah, wanna double major in math and physics.
Vertices on the y axis don't exist though. Sub in x = 0 and you get y^2 = -b^2 which unless y = b = 0, is impossible, and b=0 can't be the case either.I don't think so- what about vertices on y axis? You have to deal with a quadratic for those
I really wish I chose to pursue my hobbies (which was math) instead of doing what would be appealing to our family image.Am so glad to hear that
Sorry I messed up there- forgot this was a hyperbola not an ellipse.Vertices on the y axis don't exist though. Sub in x = 0 and you get y^2 = -b^2 which unless y = b = 0, is impossible, and b=0 can't be the case either.
Damn didn't know a could be negative only got half the solutions|a|e - |a| = |a|(e-1) = 1
Therefore |a| = 1
a= 1 or a = -1
Oh and also you can do the other vertex:
|a|e + |a| = |a|(1+e) = 1
Therefore |a| = 1/3
a = 1/3 or a = -1/3
Damn. Are you sure? Isn't 'a' theoretically defined as the length of the semi-major axis so it can't be negative?Damn didn't know a could be negative only got half the solutions
I'm not sure if they'd require that honestly, a and b are usually assumed to be positive anyway. I just did that for completeness.Damn didn't know a could be negative only got half the solutions
pls i hope you're rightDamn. Are you sure? Isn't 'a' theoretically defined as the length of the semi-major axis so it can't be negative?
I'm guessing, either I'm wrong in doing the negative, or even if you can technically do a negative they won't care about it. Like, I won't pretend to know all the definitions in conics, so you could totally be right. And in that case doing the negative is wrong.Damn. Are you sure? Isn't 'a' theoretically defined as the length of the semi-major axis so it can't be negative?
Yeah same got stuck thereHow did everyone prove theta > pi/4
I let N > 0, so theta > h/r, didn't know where to go from there
You let N>0, so tantheta>h/r, But tantheta = r/h, therefore tantheta> 1/tantheta and then you just rearrangeHow did everyone prove theta > pi/4
I let N > 0, so theta > h/r, didn't know where to go from there
I did N>=0, so sin(theta)>=h/r cos(theta)How did everyone prove theta > pi/4
I let N > 0, so theta > h/r, didn't know where to go from there
Ah shit, I was thinking in my head that tantheta = h/r for some reason. Ah well, another mark lostYou let N>0, so tantheta>h/r, But tantheta = r/h, therefore tantheta> 1/tantheta and then you just rearrange
As written in the question, a can be negative.Damn. Are you sure? Isn't 'a' theoretically defined as the length of the semi-major axis so it can't be negative?