# Calculus

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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=

• Calculus -

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 = (y-3)/(x-12)
then
(y-3)/(x-12) = -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 = 3-y
plug back into ellips
(3-y)^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

• Calculus -

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 :)

• Calculus -

Hmmm. I get 10x-15y = 75

http://www.wolframalpha.com/input/?i=plot+x^2%2B4y^2%3D36+and+y%3D3+and+10x-15y%3D75

• Calculus -

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

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