# calculus

posted by on .

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 - ,

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!