Posted by **mike** on Tuesday, October 19, 2010 at 11:19pm.

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**, Tuesday, October 19, 2010 at 11:44pm
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 = (y-3)/(x-14)

so (y-3)/(x-14) = -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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