4U Revising Game (1 Viewer)

[ronnknee]

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MONEY

Substitute a number for each letter so that the equation works.

Irrelevant

 

[conics2008]

shit sorry i forgot about this game.. hmm part i, i just did it now

you said

the point P(x0,y0) lies on the hyperbola x2 / a2 - y2 / b2 = 1

(i) Show that the acute angle @ between the asymptote satisfies

tan@ = 2ab / (a2 - b2)


use the angel formula | m1+m2 / 1-m1m1 | crap

then you get

b/a -(-b/a) / 1+ b/a . -b/a

simplfy you get

2b/a / a^2 -b^2 /a^2

simpfly more

you get 2b / a^2-b^2

therefore tan@ = 2b/a^2-b^2

ok part ii.. thanks guys

 

[conics2008]

(ii) If M and N are the feet of the perpendiculars drawn from P to the asymptotes, show that MP.NP = a2b2 / (a2 + b2)

guys what do you mean by this..

????? im getting confused by what its asking..

(ii) If M and N are the feet of the perpendiculars drawn from P to the asymptotes, show that MP.NP = a2b2 / (a2 + b2)

I can draw aline perpendicular from P to the x-axis but what does the other thingy mean..

drawn from P to the asymptotes...

 

[namburger]

(ii) If M and N are the feet of the perpendiculars drawn from P to the asymptotes, show that MP.NP = a2b2 / (a2 + b2)

guys what do you mean by this..

????? im getting confused by what its asking..

(ii) If M and N are the feet of the perpendiculars drawn from P to the asymptotes, show that MP.NP = a2b2 / (a2 + b2)

I can draw aline perpendicular from P to the x-axis but what does the other thingy mean..

drawn from P to the asymptotes...

http://i237.photobucket.com/albums/ff3/Namburger/Hyperbolaq.jpg

 

[conics2008]

^^^^^^^^^^^^ wtfff..

those chicks aren't going to solve the problem...

 

[conics2008]

thanks namburger.. i went to the x-axis and the other one to the asymp crap.. thanks alot =)

 

[YannY]

solve z^2=-4

 

[conics2008]

ok.. im stuck now..

part ii..

I know i have to use perpindcular distance formula, but the problem is at the end, im stuck with b^2/a^2+b^2 ???

any suggestion ??

 

[conics2008]

solve z^2=-4

ok!

z^2=-4 >> arg = pi

mod of z^2 = 4 ... therefore z = 2

2@ = pi + 2pik >> @ = pi+2pi k / 2 k=0,1,2,3 but we are only looking for 2 since its ^2

therefore @ = pi/2 and 3pi/2

therefore roots are z=2cispi/2 and z= 2cis3pi/2

 

[namburger]

2i, -2i

 

[Rhanoct]

(x)^0 = 1

 

[conics2008]

(x)^0 = 1

start from this x^m

rule x^m/ x^m = 1

but x^m / x^m = x^m-m = x^0

therefore x^0 =1 xD

 

[ronnknee]

solve z^2=-4

z
= ±root(-4)
= ±2i

Lol at De Moirve's theorem method

 

[conics2008]

hows your conics going? still losing at ur own game?
__________________

hmm i did part i, part ii, is confusing me... read the post i posted...


watch your tone buddy !!

 

[Mark576]

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Substitute a number for each letter so that the equation works.

Spoiler

S = 9
E = 5
N = 6
D = 7
M = 1
O = 0
R = 8
Y = 2

Win.

 

[ronnknee]

(x)^0 = 1

Is this even a question?

 

[Slidey]

x^0 does not equal 1 if x=0.

 

[namburger]

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MONEY

Substitute a number for each letter so that the equation works.

Multiple answers
9893
+
1088
10981

 

[ronnknee]

Alternatively, let everything else be 0
Anyway guys, let's stay on topic :|
I pasted all the questions to the first post so it's more accessible for new posters =)

 

[ronnknee]

I'm thinking (correct me if I'm wrong, because I didn't study 4U) that it's a permitations question:

Spoiler

7 distinct options (z, a, p, o, n, r, g)

P⁷₄
= (7!) ÷ ((7!)-(4!))
= 7! ÷ 3!
= 5040 ÷ 6
= 840

Yea it's a permutation question but it's a lot more complicated than just 7P4. You'd need to consider some cases because right now, you're not including solutions that have repetitions eg. AAAA, NONO, APAN