hscishard
Active Member
What is your final answer?
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Preliminary mathematics marathon (4 Viewers)
- Thread starter ohexploitable
- Start date
hscishard
Active Member
Sorry exploit, I got a trig question. How can you show in b i) without using the double angle forumale?
ohexploitable
hey
This marathon ain't so good cos liek most ppl haven't done as much as us, liek my school has only done basic algebra, and geometry.
ohexploitable
hey
Liek the only people who have posted is me, you, 2010ers and ppl who have already done the HSC.
mitchy_boy
blue
New question?
LM and FG are two chords of the circle at right angles to each other, and they intersect at P. From P, a line perpendicular to GM is drawn, meeting it at S. SP is produced to meet LF at T.
a) Show that ∆ LTP is isosceles.
b) Show that T is the midpoint of LF.
LM and FG are two chords of the circle at right angles to each other, and they intersect at P. From P, a line perpendicular to GM is drawn, meeting it at S. SP is produced to meet LF at T.
a) Show that ∆ LTP is isosceles.
b) Show that T is the midpoint of LF.
hscishard
Active Member
You're right.yeah your right
New question?
LM and FG are two chords of the circle at right angles to each other, and they intersect at P. From P, a line perpendicular to GM is drawn, meeting it at S. SP is produced to meet LF at T.
a) Show that ∆ LTP is isosceles.
b) Show that T is the midpoint of LF.
Let angle LFG = A
and angle FLM = B
,: A + B = 90 deg (since in triangle LPF angle LPF = 90 deg)
.: angle LMG = angle LFG = A (angles on same chord LG)
& angle FGM = angle FLM = B (angles on same chord MF)
.: angle SPM = B (angle PSM = 90 deg)
.: angle LPT = angle SPM = B (vert opp angles)
.: angle TLP = B = angle LPT in triangle LTP
.: triangle LTP is isosceles
Similarly in triangle GPS, angle GPS = A (angle PSG = 90 deg)
.: angle TPF = angle GPS = A (vert opp angles)
.: in triangle TFP, angle TPF = A = angle TFP
i.e TPF is an isosceles triangle
.: PT = TF
but TP = LT
.: PT = TF
i.e. T is mid-point of LF
QED
Edit
Didn't realise this is a marathon question; can someone pl post a question for me.
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ohexploitable
hey
.Sorry exploit, I got a trig question. How can you show in b i) without using the double angle forumale?
frenzal_dude
UTS Student
Then make it into a general quadratic expression and find the minimum value of the parabola.
What's the final answer? I got 34.45 and 35.46.
hscishard
Active Member
Hi 5!mitchy_boy
blue
Woah that was ext2 harder ext1 question. I don't have time to go through now but I think you might have hit it on the head. Nice work man.
hscishard
Active Member
Why would you post that on a prelim? Lol
That is pretty hard
That is pretty hard
ohexploitable
hey
New Question:
Differentiate:
=\sqrt[3]{(3x+5)(2x-7)^{2}})
Differentiate:
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Very tempting to do that one above using logarithmic differentiation....
Have you Prelimers done series? I have a nice series question here:
 )
Otherwise, another one from differential calculus:
The tangent to the hyperbola xy = k at the point P(x1, y1) cuts the x-axis at X and the y-axis at Y. Show that the area of triangle OXY is constant.
Have you Prelimers done series? I have a nice series question here:
Otherwise, another one from differential calculus:
The tangent to the hyperbola xy = k at the point P(x1, y1) cuts the x-axis at X and the y-axis at Y. Show that the area of triangle OXY is constant.
hscishard
Active Member
I'll try the second one.Very tempting to do that one above using logarithmic differentiation....
Have you Prelimers done series? I have a nice series question here:
Otherwise, another one from differential calculus:
The tangent to the hyperbola xy = k at the point P(x1, y1) cuts the x-axis at X and the y-axis at Y. Show that the area of triangle OXY is constant.
f'(x) = -k/x^2
Therefore tangent is
y-intercept:
x intercept
Area of triangle = 1/2AB
B is perp height.
k is a constant and points of P(x1,y1) are also constants.
Therefore area is constant?
Edit: Oh For the first one, I think S = 7/4 (calculator LOL)
Last edited:
ohexploitable
hey
I think P is any point on the curve, so not constant.
P is not a constant lol. You need to use the condition (ie. xy=k) since you are not using parametrics.I'll try the second one.
f'(x) = -k/x^2
Therefore tangent is
y-intercept:
x intercept
Area of triangle = 1/2AB
B is perp height.
k is a constant and points of P(x1,y1) are also constants.
Therefore area is constant?
And Trebla, this is a typical 4U conics question isn't it? It's a bit mean asking it to the prelimers
ohexploitable
hey
I was thinking of like a logical approach...
A = 1/2 ab
height = y
length = x
xy=k
therefore A = 1/2 k
since k is constant, area is constant
amirite?