calculus
posted by mike .
use implcit differenciation to find an eqaution of both the tangent line to the ellipse:
2x^2 + 4y^2 = 36
that passes through the points: 14,3

calculus 
Reiny
First of all, did you notice that the given point lies outside the ellipse ?
So from that exterior point there will be two different tangents.
find dy/dx ....
4x + 8y dy/dx = 0
dy/dx = 4x/(8y) = x/(2y)
let the point of contact be P(x,y)
slope of tangent = (y3)/(x14)
so (y3)/(x14) = x/(2y)
which reduces to
x^2 + 2y^2 = 14x  6y (#1)
the original could be reduced to
x^2 + 2y^2 = 18
subtract from #1
14x  6y = 18
7x  3y = 9
x = (9+3y)/7
sub back into x^2 + 2y^2 = 18
I ended up with
107y^2 + 54y  801 = 0
which actually factored to
(y+3)(107y  267) = 0
y = 3 or y = 267/107
then x = 18/7 or x = 252/107
( I cheated and used my fraction key on my calculator)
not done yet, have to find the equations ...
1st tangent:
slope: sub (3, 18/7) into x/(2y) to get
slope = 6/(36/7) = 7/6
equation is 7x  6y = c
sub in (14,3)
98  18 = c
c = 80
first tangent : 7x  6y = 80
I will let you do the second, good luck!
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