Conics question.. (1 Viewer)

chochibi

New Member
Joined
Sep 4, 2005
Messages
28
Location
sydney =P
Gender
Male
HSC
2006
hey guys.. this was in a past paper for my school and it's kinda getting to me..


ii) Prove that the tangent to the hyperbola x^2/4 - y^2/5 = 1 at the point P (2sec@ , (√5)tan@) is : (xsec@)/2 - (ytan@)/2 = 1

i got up to here and the next question kidna threw me off..

iii) This tangent cuts the asymtotes in L & M. Find the distance LM
iv) Hence Prove that the area of ∆OLM is independent of the position of P (O is the origin)

can anyone help me with this?
 

Mountain.Dew

Magician, and Lawyer.
Joined
Nov 6, 2005
Messages
825
Location
Sydney, Australia
Gender
Male
HSC
2005
chochibi said:
hey guys.. this was in a past paper for my school and it's kinda getting to me..


ii) Prove that the tangent to the hyperbola x^2/4 - y^2/5 = 1 at the point P (2sec@ , (√5)tan@) is : (xsec@)/2 - (ytan@)/2 = 1

i got up to here and the next question kidna threw me off..

iii) This tangent cuts the asymtotes in L & M. Find the distance LM
iv) Hence Prove that the area of ∆OLM is independent of the position of P (O is the origin)

can anyone help me with this?
(iii) you should know what the asymptotes of the hyperbola are. from memory, i think they are y = +/- (b/a)x ==> substitute the equation of the tangent into the SEPERATE equations of the asymptotes (ie, sub into y = - (b/a)x AND ALSO y = (b/a)x) to find the two intersecting points, L & M.

then, use the distance formula to find distance LM.

once u got distance LM, realise that OLM can be seperated into right angled triangles. draw a line from O perpendicular to LM, to meet at LM. this line's distance is found using the perpendicular distance formula.

then, area of OLM = 1/2(LM)( _|_ distance from O) = some constant <== independent of P.

hope it helps, M.D.
 

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

Top