Posted by karl on Saturday, September 5, 2009 at 5:52pm.
Show that the tangents to the curve X=ln((y^2)-6y+11) ,at the points where X=ln6,meet at the point ((ln6)-(4/3),3)
A level Mathematics - Reiny, Saturday, September 5, 2009 at 10:50pm
first of all we need dy/dx for the relation, that would be
1 = (2y(dy/dx) - 6dy/dx)/(y^2 - 6y + 11)
simplifying and solving for dy/dx gave me
dy/dx = (y^2 - 6y + 11)/(2y-6)
then let's sub in x=ln6 into the original
ln6 = ln(y^2 - 6y + 11)
6 = y^2 - 6y + 11
solving as a quadratic, it factored, I got
y = 1 or y = 5
so at the point (ln6,1), dy/dx = -3/2
and at point (ln6,5), dy/dx = +3/2
y-1 = (-3/2)(x-ln6) which reduced to
3x + 2y = 2 + 3ln6 (#1)
y-5 = (3/2)(x-ln6) gave me
3x - 2y = 3ln6 - 10 (#2)
adding #1 and #2 gave me
x = ln6 - 4/3
and carefully subbing that back into #1 gave me y = 3
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