Find equations of both tangent

Answer: y = 3 and 3x + 8y - 30 = 0

Let the equation be y - 3 = m(x - 2) => y = m(x - 2) + 3
Sub. y = m(x - 2) + 3 into x^2 + 4y^2 = 36
x^2 + 4[m(x - 2) + 3]^2 = 36
x^2 + 4m^2 (x - 2)^2 + 24m(x - 2) + 36 = 36
x^2 + 4m^2 x^2 - 16m^2 x + 16m^2 + 24mx - 48m + 36 = 36
(1 + 4m^2) x^2 + (24m - 16m^2) x + 16m^2 - 48m = 0

since the tangents cut the ellipse at one point,
(24m - 16m^2)^2 - 4(1 + 4m^2)(16m^2 - 48m) = 0
(12m - 8m^2)^2 - (1 + 4m^2)(16m^2 - 48m) = 0
(64m^4 - 192m^3 + 144m^2) - (64m^4 - 192m^3 + 16m^2 - 48m) = 0
128m^2 + 48m = 0
m = 0 or m = -3/8

thus, the equation of tangent are y - 3 = 0 and y - 3 = (-3/8)(x - 2)
i.e. y = 3 and 3x + 8y - 30 = 0


2013-05-21 11:16:36 補充：
Methos 2

Let the intersection point be (a, b)
differentiate both sides of x^2 + 4y^2 = 36 w.r.t. x
2x + 8y(dy/dx) = 0 => dy/dx = -x/(4y)

2013-05-21 11:16:57 補充：
Solve -a/(4b) = (b - 3)/(a - 2) ------ (i) and a^2 + 4b^2 = 36 ------ (ii)
from (i), -a^2 + 2a = 4b^2 - 12b => a^2 + 4b^2 - 2a - 12b = 0 ------ (iii)
(iii) - (ii), a = 18 - 6b ------ (iv)
Sub. (iv) into (ii), (18 - 6b)^2 + 4b^2 = 36
10b^2 - 54b + 72 = 0
b = 3 or 12/5

2013-05-21 11:17:56 補充：
when b = 3, a = 0, dy/dx = 0, the equation of tangent is y = 3
when b = 12/5, a = 18/5, dy/dx = -3/8,
the equation of tangent is y - 12/5 = (-3/8)(x - 18/5) => 3x + 8y - 30 = 0