MedVision ad

Help with random 3U questions! (1 Viewer)

WildKoala

New Member
Joined
Mar 28, 2012
Messages
19
Gender
Female
HSC
N/A
If you can answer any of the following (w/ working out would be helpful)!

1. Expand and simplify (a+b)^n + (a-b)^n


2.Let A be the point (-1,0) on the unit circle and suppose that the line throagh A with gradient t meet the circle at P in the first quadrant. OP and AP make angles of theta and alpha respectively with the positive x-axis.
i) what range of values can t take?​
ii) Prove P has x-coordinate 1/t^2/1+t^2 and find the y-cord of P​
iii) Deduce that tan theta = 2t/ 1-t^2, for t cannot equal 1​
iv) Find the value of t if alpha = 22.5 degrees​


3. A cardioid curve is defined by the following pair of parametric eqn:
x = 2costheta - cos2theta
y= 2sintheta - sin2theta
for 0 < theta < 2pi

The cardioid curve and the vertical line x=1 intersect in the 1st quadrant at a point P. find the coordinates of the point P
 

cineti970128

Member
Joined
Apr 25, 2012
Messages
139
Gender
Male
HSC
2014
Q1
(a+b)^n + (a-b)^n
note that their is (-) sign hence if you put them in general form you should see that every second tern ie for r= odd, the term cancels each other. Whereas for r= even ,they add up

Hence = 2 (C0 a^n + C2 a^(n-2)b^2 + C4 a^(n-4) b^4 + ..... )
the last tern depends on the nature of n
 

cineti970128

Member
Joined
Apr 25, 2012
Messages
139
Gender
Male
HSC
2014
2 i)
with this question draw a diagram to assist you in understanding the question

Note that the line should intersect the circle in the FIRST QUANDRANT
ie maximum t = when P lies on (0,1)
hence 0 < t < 1 (greater or equal sign)
 

cineti970128

Member
Joined
Apr 25, 2012
Messages
139
Gender
Male
HSC
2014
ii ) I will use algebraic method in approaching this question
1. you can make two equations
y = tx + t
y^2 + x^2 = 1
Hence solving simultaneously
x = (1-t^2)/(1+t^2)
hence subing this to linear equation
y = (2t/(1+t^2))

iii) hence tan (A) = y/x
= (2t/(1-t^2))

iv) using circle geometry angle by OP = 2 X angle by AP
(angle at the centre)
Hence theater is 45 hence tanA =1
using iii)
you can produce t^2 + 2t - 1
solving using quadratic formula t = -1+or root 2
using i) t cannot be negative
hence t = -1 + (2)^1/2
 

cineti970128

Member
Joined
Apr 25, 2012
Messages
139
Gender
Male
HSC
2014
3
x = 2costheta - cos2theta

in this equation put x = 1 and let cos theata be u
then using compound angle
u(u-1) = 0
solving it
cosA = 0 or cos A= 1
A = 0, pi/2, 3pi/2, or 2pi
hence
y = 0, 2 , -2, 0
hence coordinates are (1,0) and (1, 2)
(1, -2) is rejected
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top