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December 19, 2014

December 19, 2014

Posted by **EYOB** on Sunday, March 7, 2010 at 2:52am.

- math -
**Reiny**, Sunday, March 7, 2010 at 7:39amNon-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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