Calculus
posted by Tim on .
Find equations of both the tangent lines to the ellipse x^2 + 4y^2 = 36 that pass through the point (12, 3).
smaller slope y=
larger slope y=

the the point of contact be P(x,y)
Make a sketch to see that there would be two points P
2x + 8y dy/dx = 0
dy/dx = 2x/(8y) = x/(4y)
slope of tangent = (y3)/(x12)
then
(y3)/(x12) = x/(4y)
4y^2  12y = x^2 + 12x
x^2 + 4y^2 = 12x + 12y
but x^2 + 4y^2 = 36
12x + 12y = 36
x + y = 3
x = 3y
plug back into ellips
(3y)^2 + 4y^2 = 36
9  6y + y^2 + 4y^2 = 36
5y^2  6y  27 = 0
(y  3)(5y + 9)
y = 3 or y = 9/5
then
x = 0 or x = 3  (9/5) = 24/5
so P could be (0,3) or P could be (24/5, 9/5)
case 1: P is (0,3)
then slope = x/(4y) = 0/12 = 0
ahhh, a horizontal line
y = 3
case 2 :P is (24/5, 9/5)
slope = (24/5) / (4(9/5)) = 2/3
so (y + 9/5) = (2/3)(x  24/5)
3(y + 9/5) = 2(x  24/5)
3y + 27/5 = 2x  48/5
3y  2x = 21/5
15y  10x = 21
10x  15y = 21
check my answer for reasonableness by looking at your graph 
You can see that tangent from (0,3), the top of the ellipse, to (12,3) is y = 3 by inspection. The other tangent requires some work :)

Hmmm. I get 10x15y = 75
http://www.wolframalpha.com/input/?i=plot+x^2%2B4y^2%3D36+and+y%3D3+and+10x15y%3D75 
Steve is right, I made a silly adding vs subtracting error in the 3rd last line
3y + 27/5 = 2x  48/5
3y  2x = 21/5
should be
3y + 27/5 = 2x  48/5
3y  2x = 75/5
to give
10x  15y = 75 , reducing to
2x  3y = 15