For what values of a and b is the line 2x + y = b tangent to the parabola y = ax^2 when x = 2?

Question

For what values of a and b is the line 2x + y = b tangent to the parabola y = ax² when x = 2?

 

 

ANSWER: a = -1/2, b = 2

#73, p183

Answer 1

If 2x+y=b, then y = -2x+b  So, the slope of the line is -2.

 

If y = ax², then y' = 2ax

 

The slope of the tangent line to the parabola at x = 2 is 2a(2) = 4a.

 

So, for the line to be tangent to the parabola at x = 2, we want the slope of the line to equal 4a

 

4a = -2   So, a = -1/2

 

The equation of the parabola is therefore y = (-1/2)x²

 

The line intersects the parabola when x = 2 and y = (-1/2)(2)² = -2

 

The equation of the line is y = -2x + b

 

So, at the point (2,-2), we have -2 = -2(2)+b

 

Therefore, b = 2

 

 

Answer 2

1. y = -2x + b  (first equation written in point slope)

2. y = ax²    (2nd equation)

3. y' = 2ax  (1st derivative of the parabola's function--gives us formula for slopes of tangent lines to the parabola)

 

Now subsitute x=2 in each of the above to see where it leads.

 

1. y = -4 + b

2. y = 4a

3. y' = 4a (so the slope of the tangent line at x=2 is 4a).

 

Substitute 4a from equation 2 into the first equation for y, giving:

 

4. 4a = -4 + b or 4a - b = -4.

 

At this point we have one equation containing a and b. We need another.

 

Let's see of we can make use of the fact that the slope of the tangent line is 4a.

 

Can we find two different points that the tangent line passes through?

 

From 1, the y-intercept is (0,b).

From 2, when x=2, y=4a, so the tangent line passes through (2,4a).

 

The slope of a line containing those two points is:

 

(4a - b)/(2-0)

 

But that slope from 3 is 4a.

 

So (4a - b)/2 = 4a => 4a - b = 8a => 4a + b = 0 (call that equation 5)

 

Repeating equations 4 and 5.

 

4. 4a - b = -4

5. 4a + b = 0

 

Using addition method eliminates b.

 

8a = -4

a = -1/2

b = 2 (substitute -1/2 for a in 4).

 

See https://www.desmos.com/calculator/itnddskoxk