Question

hello

Please solve my problem.....

if the line y = mx + 1 is tangent to parabola y(square) = 4x then find the value of m .

Answer 1

Hi

Please see the attached file !

Answer 2

Given Parabola is y^2 = 4x  ---------------- (1)

Given line is y = mx + 1  --------------- (2)

Substitute (2) in (1), we get

(mx + 1)^2 = 4x

m^2x^2 + 1 + 2mx = 4x

m^2x^2 + 1 + 2mx - 4x = 0

m^2x^2 + x(2m - 4) + 1 = 0

The tangent line touches the parabola, they have one point in common.The Quadratic Equations have only one solution When the discriminant is 0, Then

D = b^2 - 4ac = 0

    = (2m - 4)^2 - 4m^2 = 0

    = (2m - 4)^2 - (2m)^2 = 0

    = ((2m - 4) + 2m)(2m - 4) - 2m) = 0

    = (2m - 4 + 2m)(2m - 4 - 2m) = 0

    = (4m - 4)(-4) = 0

    = 4m - 4 = 0
 
    = 4m = 4

    m = 4/4

    m = 1.



Hope this helps!