F fullonoob fail engrish? unpossible! Joined Jul 19, 2008 Messages 465 Gender Male HSC 2010 Mar 30, 2010 #1 The equation of the tangent to y=xlogx at x=e is y = 2x-e. Find the distance from this tangent to the origin
The equation of the tangent to y=xlogx at x=e is y = 2x-e. Find the distance from this tangent to the origin
R Rezen Member Joined Mar 12, 2009 Messages 62 Gender Male HSC 2010 Mar 30, 2010 #2 perpendicular distance?
alcalder Just ask for help Joined Jun 26, 2006 Messages 601 Location Sydney Gender Female HSC N/A Mar 30, 2010 #3 fullonoob said: The equation of the tangent to y=xlogx at x=e is y = 2x-e. Find the distance from this tangent to the origin Click to expand... Perpendicular distance, IS the shortest distance. Line perpendicular to y = 2x - e has the gradient -1/2. It goes through (0,0) therefore has the equation y =-x/2 These two lines cross at -x/2 = 2x-e -x = 4x - 2e x = 2e/5 y = -e/5 distance = sqrt((-e/5)2 + (2e/5)2)
fullonoob said: The equation of the tangent to y=xlogx at x=e is y = 2x-e. Find the distance from this tangent to the origin Click to expand... Perpendicular distance, IS the shortest distance. Line perpendicular to y = 2x - e has the gradient -1/2. It goes through (0,0) therefore has the equation y =-x/2 These two lines cross at -x/2 = 2x-e -x = 4x - 2e x = 2e/5 y = -e/5 distance = sqrt((-e/5)2 + (2e/5)2)