math
posted by EYOB on .
if the line y=2x+b and the parabola y^2=8(x+2) meet exactly at one point,then what is the value of b

NonCalculus way:
Let's find their intersection, ...
(2x+b)^2 = 8(x+2)
4x^2 + 4bx + b^2  8x  16 = 0
4x^2 + (4b8)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
(4b8)^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.