Posted by EYOB on Sunday, March 7, 2010 at 2:52am.
Non-Calculus way:
Let's find their intersection, ...
(2x+b)^2 = 8(x+2)
4x^2 + 4bx + b^2 - 8x - 16 = 0
4x^2 + (4b-8)x + b^2 - 16 = 0
this is a quadratic with
A = 4
B = 4b -8
C = b^2 - 16
IF this is to have only one root, then the B^2 - 4AC, or the discriminant, must be zero.
so
(4b-8)^2 - 4(4)(b^2 - 16) = 0
16b^2 - 64b + 64 - 16b^2 + 256 = 0
64b = 320
b =5
Using Calculus:
for y^2= 8x + 16
2y(dy/dx) = 8
dy/dx = 8/2y ----> slope of the tangent
but we know the slope of the tangent is 2 from y - 2x+b
then 8/2y = 2
4y = 8, then y = 2
sub that into parabola
4 = 8x + 16
x = -12/8 =- 3/2
so the point of contact is (- 3/2,2) which must satisfy
y = 2x + b
2 = 2(-3/2) + b
2 = -3 + b
b = 5 , just like before.
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