A level Mathematics
posted by karl on .
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)

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)/(2y6)
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
first tangent:
y1 = (3/2)(xln6) which reduced to
3x + 2y = 2 + 3ln6 (#1)
second tangent:
y5 = (3/2)(xln6) 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
YES!!!!!