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

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


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
k=3%2F4

The difference between the two equations should be 0.
y%2F2-k%2F2+-+y%5E2%2F6=0
.
.


graph%28400%2C400%2C-3%2C3%2C-3%2C3%2C-sqrt%286x%29%2Csqrt%286x%29%2C2x%2B3%2F4%29

Answer by ikleyn(52781) About Me  (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?
~~~~~~~~~~~~~~~~~~~~~ 

The key idea of solving this problem is that the system

2x - y + k = 0,     (1)
y%5E2 = 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):

y%5E2 = 3*(2x),   or

y%5E2 = 3*(y-k),

y%5E2+-+3y+%2B+3k = 0.


The discriminant of this quadratic equation is 

d = 3%5E2+-+4%2A3k = 9 - 12k.


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

9 - 12k = 0,

which gives you k = 9%2F12 = 3%2F4.

Answer. k = 3%2F4.




Parabola y%5E2 = 6x and the straight line 2x - y + 3%2F4 = 0.


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



Answer by Alan3354(69443) About Me  (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