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

#### calamebe

##### Active Member
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

Last edited:
• BLIT2014

#### calamebe

##### Active Member
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

##### Member
|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

##### Member
Yeah, wanna double major in math and physics.
Am so glad to hear that

• calamebe

#### calamebe

##### Active Member
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

##### Member
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.

• calamebe

#### shervos

##### Member
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

##### Member
But my other point still stands

#### pikachu975

##### I love trials
Moderator
|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

##### New Member
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

##### Active Member
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

##### New Member
How did everyone prove theta > pi/4

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

• pikachu975

#### kibellion

##### New Member
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

##### Active Member
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

##### I love trials
Moderator
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

##### New Member
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

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

#### Green Yoda

##### Hi Φ
Moderator
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

##### New Member
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

##### Well-Known Member
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).