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)

Answer 1

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

first tangent:
y-1 = (-3/2)(x-ln6) which reduced to
3x + 2y = 2 + 3ln6 (#1)

second tangent:
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

YES!!!!!