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

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!!!!!

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