VCE Specialist 3/4 Question Thread!

[barydos]

OK, so ABD forms an isosceles triangle if you draw a chord between B and D. After that, you'll see angle ADB is also x, and then by utilising the tangent/chord angle in the opposite segment rule, you'll also see angle BCD is x.

For part b, you'll need to use the cross over of two secants rule so that you get y^2=a(a+b).

for part a... ahh i see thanks,
but for part b, im not actually aware of this rule, could you possibly enlighten me on this, or is there another method? thanks

[Stick]

http://www.sparknotes.com/math/geometry2/theorems/section5.rhtml

Theorem 3 on that page.

[barydos]

http://www.sparknotes.com/math/geometry2/theorems/section5.rhtml

Theorem 3 on that page.

wow wtf, that was totally not in chapter 1 of the textbook.... (unless i missed it haha)
but thanks a lot man, that was really interesting and helpful



On another note, for this question... (solving for x and y)
Can I work it out by expanding (x+yi)^2, and then equating coefficients then simultaneously solving for x and y?
Is there another method for it?

[Stick]

Why not square root both sides? Remember there will be two solutions!

[barydos]

Why not square root both sides? Remember there will be two solutions!

Wow, you're amazing. FULLY APPRECIATE IT DUDE
How did I miss that! haha

[barydos]

Gah, I was hoping I didn't have to ask another question for the day, but I guess I had to!



Part a) I believe is just 90 deg because the pythagorean triad 9-12-15...
Part b) i just dont know :O

[#1procrastinator]

How do you show that has an asymptote at y=1? No information is given about a or b.

It's a MC (asks for all the asymptotes of the above function) and the options are:
A) x=a and x=-a
B)x=b and x=-b
C) y=1 and x=a
D) y=0, x=a and x=-a
E) y=1, x=a and x=-a

[BubbleWrapMan]

Perform the polynomial division, you should get 1 + (a^2-b^2)/(x^2-a^2). The numerator is constant and the denominator approaches infinity as x does, so the whole expression approaches 1 + 0 = 1.

[#1procrastinator]

^ Thanks. Damn I gotta learn how to divide polynomials lol

[bonappler]

I'm having some trouble with "regions in the complex plane," could someone tell me the thinking processes you have to go through for this:
Let S={z:2mod(z)≤ mod(z-3+3sqrt3i)}
Sketch S

[e^1]

Using Essentials textbook, but I'm stuck understanding part of the section.

If anyone could, then please explain

edit: Thanks pi, appreciate it

[pi]

Using similar triangles:
OC/OB = OC'/OB'
Hence, OC' = (OC)(OB')/OB
Hence, cos(theta)= (OC)(OB')/OB
As OB' = 1
cos(theta)= OC/OB

You can do a similar one for sine, hope that helps

[barydos]

Anyone know how to get around part b) (both i and ii)?

[brightsky]

b) i) You must first find the point on the circumference of the circle which is nearest the origin. There are a number of ways to find this point. I think the quickest way involves first finding the equation of the line connecting centre (-2,2sqrt(3)) and the origin (0,0), and then finding point on the circumference of the circle (nearest the origin) through which the line passes by solving the cartesian equation of the circle and the equation of the line simultaneously. After you find said point, the question becomes trivial.

ii) The required z corresponds to the point (-2, 2sqrt(3) - 2). Simply find the argument of this point.

Edit: Derped.

[barydos]

b) i) You must first find the point on the circumference of the circle which is nearest the origin. There are a number of ways to find this point. I think the quickest way involves first finding the equation of the line connecting centre (-2,2sqrt(3)) and the origin (0,0), and then finding point on the circumference of the circle (nearest the origin) through which the line passes by solving the cartesian equation of the circle and the equation of the line simultaneously. After you find said point, the question becomes trivial.

ii) The required z corresponds to the point (-2, 2sqrt(3) - 2). Simply find the argument of this point.

Hey thanks a lot brightsky, that helps heaps

EDIT: Actually, for part b) ii)

with the point (-2, 2sqrt(3)-2),
finding the argument.... I'm doing
tan theta = (2sqrt3 -  2)/(2) = sqrt3 - 1
but  I can't find theta this way... I'm sure I messed up somewhere... but I can't find it D: