2017 HSC Mathematics Extension 2 paper thoughts? (1 Viewer)

[calamebe]

Can someone please write a solution for 16b

|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

 

[calamebe]

Oh wow, I guess I'll retract my previous statement then.
Good on you either way - do you plan on pursuing mathematics in university?

Yeah, wanna double major in math and physics.

 

[shervos]

|a|e - |a| = |a|(e-1) = 1

Therefore |a| = 1

a= 1 or a = -1

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

 

[shervos]

Yeah, wanna double major in math and physics.

Am so glad to hear that

 

[calamebe]

I don't think so- what about vertices on y axis? You have to deal with a quadratic for those

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.

 

[shervos]

Am so glad to hear that

I really wish I chose to pursue my hobbies (which was math) instead of doing what would be appealing to our family image.

 

[shervos]

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.

Sorry I messed up there- forgot this was a hyperbola not an ellipse.

 

[shervos]

But my other point still stands

 

[pikachu975]

|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 didn't know a could be negative only got half the solutions

 

[TheZhangarang]

Damn didn't know a could be negative only got half the solutions

Damn. Are you sure? Isn't 'a' theoretically defined as the length of the semi-major axis so it can't be negative?

 

[calamebe]

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.

 

[TheZhangarang]

How did everyone prove theta > pi/4

I let N > 0, so theta > h/r, didn't know where to go from there

 

[kibellion]

Damn. Are you sure? Isn't 'a' theoretically defined as the length of the semi-major axis so it can't be negative?

pls i hope you're right

 

[calamebe]

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 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]

How did everyone prove theta > pi/4

I let N > 0, so theta > h/r, didn't know where to go from there

Yeah same got stuck there

 

[MaxATARLoading]

How 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 rearrange

 

[MaxATARLoading]

Can someone post the answers for Questions 8, 9 and 10 in MCQ? I got B, D and C respectively

 

[Green Yoda]

How 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)
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]

You let N>0, so tantheta>h/r, But tantheta = r/h, therefore tantheta> 1/tantheta and then you just rearrange

Ah shit, I was thinking in my head that tantheta = h/r for some reason. Ah well, another mark lost

 

[InteGrand]

Damn. Are you sure? Isn't 'a' theoretically defined as the length of the semi-major axis so it can't be negative?

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).