Posted by karl on Saturday, September 5, 2009 at 5:52pm.
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!!!!!
Related Questions
maths - Show that the tangents to the curve X=ln((y^2)-6y+11) ,at the points ...
math - Find the equation of the tangent line to y=6^((x^2)-4x+8)) at x=3. I have...
12th Grade Calculus - 1. a.) Find an equation for the line perpendicular to the ...
calculus - there are two tangents lines to the curve f(x) = 3x^2 that pass ...
math - consider the curve defined by the equation y=a(x^2)+bx+c. Take a point(h,...
Math - Show that the tangent to the curve 25x^5+5x^4-45x^3-5x^2+2x+6y-24=0 at ...
Ap calc.. Dying!! Please help! - Given the curve x^2-xy+y^2=9 A) write a general...
Math - ln24 - ln6
calculus - determine the point of intersection of the tangents at the points of ...
calculus - Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show ...
For Further Reading
