L lollol New Member Joined Apr 9, 2006 Messages 14 Gender Male HSC 2007 Oct 27, 2006 #1 tangents from the pt P(m,n) touch the parabola x^2=8y at the pts A and B. show that the x-coordinates of A and B are the roots of the quadratic equation x^2-2mx+8n=0
tangents from the pt P(m,n) touch the parabola x^2=8y at the pts A and B. show that the x-coordinates of A and B are the roots of the quadratic equation x^2-2mx+8n=0
R Riviet . Joined Oct 11, 2005 Messages 5,584 Gender Undisclosed HSC N/A Oct 27, 2006 #2 Substitute P into chord of contact: xx0 = 2a(y + yo) and note that a=2 since x2=4ay mx=4(y+n) y=mx/4 - n But y=x2/8 .'. mx/4 - n=x2/8 Then multiply by 8 and bring everything to RHS to get required result. Last edited: Oct 27, 2006
Substitute P into chord of contact: xx0 = 2a(y + yo) and note that a=2 since x2=4ay mx=4(y+n) y=mx/4 - n But y=x2/8 .'. mx/4 - n=x2/8 Then multiply by 8 and bring everything to RHS to get required result.
S SoulSearcher Active Member Joined Oct 13, 2005 Messages 6,751 Location Entangled in the fabric of space-time ... Gender Male HSC 2007 Oct 27, 2006 #3 Find the equation of the chord of contact: xx0 = 2a(y+y0, a = 2, P(m,n) mx = 4(y+n) y = (mx-4n)/4 Sub into the equation x2=8y to find x-coordinates x2 = 8[(mx-4n)/4) x2 = 2mx - 8n x2 - 2mx + 8n = 0 Therefore x-coordinates of A and B are the roots of the quadratic equation x2[/sup-2mx+8n = 0
Find the equation of the chord of contact: xx0 = 2a(y+y0, a = 2, P(m,n) mx = 4(y+n) y = (mx-4n)/4 Sub into the equation x2=8y to find x-coordinates x2 = 8[(mx-4n)/4) x2 = 2mx - 8n x2 - 2mx + 8n = 0 Therefore x-coordinates of A and B are the roots of the quadratic equation x2[/sup-2mx+8n = 0
M Mumma Member Joined May 19, 2004 Messages 586 Location Sydney Gender Male HSC 2006 Oct 28, 2006 #4 You dont need equation for chord of contact Simply, dy/dx = x/4 tangent at point (t,t^2/8) y - t^2/8 = t/4(x-t) 8y - t^2 = 2t(x-t) goes through (m,n) 8n - t^2 = 2tm - 2t^2 t^2 - 2tm + 8n = 0 Since t is the x coordinate, x^2 - 2mx + 8n = 0 has roots that are the x-coordinates of A,B
You dont need equation for chord of contact Simply, dy/dx = x/4 tangent at point (t,t^2/8) y - t^2/8 = t/4(x-t) 8y - t^2 = 2t(x-t) goes through (m,n) 8n - t^2 = 2tm - 2t^2 t^2 - 2tm + 8n = 0 Since t is the x coordinate, x^2 - 2mx + 8n = 0 has roots that are the x-coordinates of A,B
airie airie <3 avatars :) Joined Nov 4, 2005 Messages 1,136 Location in my nest :) Gender Female HSC 2007 Oct 28, 2006 #5 Eww. Calculus. <.< I think I like "elementary methods" better XP
S SoulSearcher Active Member Joined Oct 13, 2005 Messages 6,751 Location Entangled in the fabric of space-time ... Gender Male HSC 2007 Oct 28, 2006 #6 airie said: Eww. Calculus. <.< I think I like "elementary methods" better XP Click to expand... I concur
airie said: Eww. Calculus. <.< I think I like "elementary methods" better XP Click to expand... I concur