Mathematics Help

Question

If the tangent line of parabola y= 4x-x^2 at M(1,3) is also the tangent line of parabola y=x^2-6x+k, then the value of 5- square root(k-1) is...

 

a.0

b.1

c.2

d.3

e.4

Answer 1

The first thing we need to do is find the equation of the tangent line.  I don't know if you know calculus, so I'll assume you don't.

 

Before we know anything, the equation of the tangent line is y = mx + b.  And the equation of the first parabola is y = 4x - x².  We know that M is a point on both the line and the parabola, so it should be a solution to both equations.  So let's solve those two equations together and see what we get.

 

y = 4x - x²

y = mx + b

 

So...

 

4x - x² = mx + b

0 = x² + mx - 4x + b = x² + x(m - 4) + b

 

But remember, we know something about the relationship between m and b, because we know that M satisfies the equation y = mx + b:

 

3 = m(1) + b

3 = m + b

b = 3 - m

 

So substitute 3 - m for b in the quadratic above:

 

x² + x(m - 4) + b = 0

x² + x(m - 4) + 3 - m = 0

 

Now solve this using the quadratic formula:

 

x = -(m-4) ± √[(m - 4)² - 4(1)(3 - m)]

      -----------------------------------------

                     2(1)

 

x = -m + 4 ± √[(m² - 8m + 16) - (12 + 4m)]

      -----------------------------------------------

                       2

 

x = - m + 4 ± √(m² - 4m + 4)

      -------------------------------

                      2

 

But wait.  Since we are solving for the intersection between a parabola and a TANGENT line, there should only be one solution.  In order for there to be only one solution, the expression inside the square root (the discriminant) has to be 0.  So that means that we want m² - 4m + 4 to be 0...

 

m² - 4m + 4 = 0

(m - 2)(m - 2) = 0

m = 2

 

That's great!  We get a single solution for m, the slope of the tangent line, which is exactly what we'd expect, because there is only one tangent line at each point.  So go back to the equation for the tangent line:

 

y = mx + b

3 = 2 * 1 + b

3 = 2 + b

b = 1

 

So the equation of the tangent line is y = 2x + 1.

 

Now let's solve that equation along with the OTHER parabola's equation:

 

y = 2x + 1

y = x² - 6x + k

 

2x + 1 = x² - 6x + k

0 = x² - 8x + k - 1

 

Solve for x:

 

x = -(-8) ± √[8² - 4(1) (k - 1)]

     -------------------------------

                  2(1)

 

x = [8 ± √[64 - 4k + 4)] / 2

x = [8 ± √[68 - 4k)] / 2

 

But again, we know that the discriminant has to be 0 because there has to be only one point where the tangent line intersects the parabola.  So 68 - 4k has to equal 0.

 

68 - 4k = 0

68 = 4k

17 = k

 

Now let's solve the problem they asked:

 

5- square root(k-1) = 5 - √(17 - 1) = 5 - √16 = 5 - 4 = 1.

 

So the answer is b.

 

With calculus you'd be able to find the slope of the tangent line more easily.  But even aside from that, there may be an easier way to do this.