SOLUTION: For what value of k the line 2x-y+k=0 is tangent to the parabola y^2=6x?

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Question 1083792: For what value of k the line 2x-y+k=0 is tangent to the parabola y^2=6x?
Found 3 solutions by josgarithmetic, ikleyn, Alan3354:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!


The difference between the two equations should be 0.

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Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
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For what value of k the line 2x-y+k=0 is tangent to the parabola y^2=6x?
~~~~~~~~~~~~~~~~~~~~~

The key idea of solving this problem is that the system

2x - y + k = 0,     (1)
 = 6x      (2)

must have one and only one solution.


    (Then the solution will be the tangent point).

From equation (1), express 2x = y - k and substitute it into equation (2):

 = 3*(2x),   or

 = 3*(y-k),

 = 0.


The discriminant of this quadratic equation is 

d =  = 9 - 12k.


The condition of having the unique solution is d = 0,   or

9 - 12k = 0,

which gives you k =  = .

Answer. k = .




Parabola = 6x and the straight line 2x - y + = 0.


Ignore writing by "josgarithmetic" since his conception and explanation are WRONG.



Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
For what value of k the line 2x-y+k=0 is tangent to the parabola y^2=6x?
--------
2y*dy = 6*dx
dy/dx = 3/y = the slope of the parabola
----
The slope of 2x-y+k=0 is 2
---
3/y = 2
y = 3/2; x = 3/8
The tangent point is (3/8,3/2)
--------
2x-y+k=0
3/4 - 3/2 + k = 0
k = 3/4

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