Hyperpola (1 Viewer)

[dawso]

hello peoples, im stuck on this question, if anyone has any ideas/solution, please help:

The tangent to the hyperbola x^2/4 - y^2=1 at the point P(2secA, tanA) meets the asymptotes at M and N. Prove that P is the midpoint of MN.

Thanks peoples
/dawso

 

[Affinity]

the bash:
the asymtyopes are:

y = x/2 and y = -x/2

the gradient at p is 1/(2sinA)

the tangent will therefore be

(x-2secA)/(2sinA) = y - tan(A)

x - 2secA = 2ysinA - 2*(sinA)^2/cos(A)

x - 2[(1-(sinA)^2)/(cosA)] = 2ysinA

x - 2cosA = 2ysinA

solving this for x simultaneously with each of the 2 asymtopes

with y = x/2
x - 2cosA = -xsinA
x = 2cosA/(1+sinA)

the other y = -x/2 gives
x = 2cosA/(1-sinA)

the mean of these 2 hence the x coordinate of the midpoint is

(1/2)*[ [2cosA(1+sinA) + 2cosA(1-sinA)]/[1-(sinA)^2] ]
= 2/cosA = 2secA = x coordinate of P

since P is on the line, it must then be the midpoint

 

[dawso]

thanks heaps, btw, i have to ask this, have you done this question before, or do u really have no life, and just enjoy solving other peoples problems, not that im complaining, i think it is sweet, but yeah

 

[Affinity]

Hmm.. none of above.. just have a habit of checking up message boards before I sleep

 

[nike33]

thats pretty harsh...especially when he helped you

you can have a so-called 'life' and still enjoy maths...

i dont get the "have you done this question before" as its pretty basic / standard so its not like it would take someone like affinity much time anyway

 

[Vampire]

lol
Affinity, dont take offence at dawso - if you knew him you would see how ironical his statement is.