Posted by **Tim** on Friday, March 14, 2014 at 5:24pm.

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 -
**Reiny**, Friday, March 14, 2014 at 7:14pm
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 -
**Damon**, Friday, March 14, 2014 at 7:48pm
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 -
**Steve**, Friday, March 14, 2014 at 9:14pm
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 -
**Reiny**, Friday, March 14, 2014 at 10:29pm
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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