|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
Last edited:
2017 HSC Mathematics Extension 2 paper thoughts? (3 Viewers)
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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
Yeah, wanna double major in math and physics.
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
Am so glad to hear that
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 did forget the other case though which I've fixed up now.
I really wish I chose to pursue my hobbies (which was math) instead of doing what would be appealing to our family image.
Sorry I messed up there- forgot this was a hyperbola not an ellipse.
pikachu975
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Damn didn't know a could be negative only got half the solutions
TheZhangarang
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Damn. Are you sure? Isn't 'a' theoretically defined as the length of the semi-major axis so it can't be negative?
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.
TheZhangarang
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How did everyone prove theta > pi/4
I let N > 0, so theta > h/r, didn't know where to go from there
I let N > 0, so theta > h/r, didn't know where to go from there
pls i hope you're right
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.
pikachu975
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Yeah same got stuck there
MaxATARLoading
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You let N>0, so tantheta>h/r, But tantheta = r/h, therefore tantheta> 1/tantheta and then you just rearrange
MaxATARLoading
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Can someone post the answers for Questions 8, 9 and 10 in MCQ? I got B, D and C respectively
Green Yoda
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I did N>=0, so sin(theta)>=h/r cos(theta)
h/r=cot(theta)
sin(theta)>=cos^2(theta)/sin(theta)
sin^2(theta)>=cos^2(theta)
sin(theta)>=cos(theta)
theta>=pi/4
Idk if it's correct but that's what I did.
TheZhangarang
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Ah shit, I was thinking in my head that tantheta = h/r for some reason. Ah well, another mark lost
As written in the question, a can be negative.
However, once you get the positive solutions for a, getting the rest of the solutions is trivial since you just negate the positive ones.
Since the question was only 2 marks, in my opinion, they should just give the full marks if you get the correct positive values (since getting the negative values isn't really the hard part or point of the question).