For what values of a and b is the line
4x + y = b
tangent to the parabola
y = ax2
when x = 5?
That doesn't help at all
To determine the values of a and b for which the line 4x + y = b is tangent to the parabola y = ax^2 at x = 5, we need to find the equation of the tangent line and then solve for the values of a and b.
Step 1: Find the derivative of the parabola y = ax^2.
The derivative of y = ax^2 is given by dy/dx = 2ax.
Step 2: Evaluate the derivative at x = 5 to find the slope of the tangent line.
Substituting x = 5 into dy/dx = 2ax, we get:
dy/dx = 2a * 5 = 10a.
So, the slope of the tangent line at x = 5 is 10a.
Step 3: Substitute the coordinates of the point (5, 25a) into the equation of the line 4x + y = b.
Substituting x = 5 and y = 25a, we get:
4(5) + 25a = b,
20 + 25a = b.
Step 4: Set the slope of the tangent line equal to the slope of the line.
Since the line 4x + y = b is tangent to the parabola, the slopes of both lines are equal. Therefore, we have:
10a = 4.
Step 5: Solve for a.
Dividing both sides of the equation by 10, we get:
a = 4/10 = 2/5.
Step 6: Substitute the value of a back into the equation from Step 3 to solve for b.
Using a = 2/5 in the equation 20 + 25a = b, we have:
20 + 25(2/5) = b,
20 + 10 = b,
b = 30.
Therefore, the line 4x + y = 30 is tangent to the parabola y = (2/5)x^2 when x = 5.
To find the values of a and b that make the line tangent to the parabola, we need to determine the point of tangency.
First, let's find the derivative of the parabola equation y = ax^2 with respect to x to find the slope of the tangent line at any given point. The derivative of ax^2 is 2ax.
Let's find the equation of the tangent line when x = 5. Substituting x = 5 into the parabola equation, we get y = a(5)^2 = 25a.
Thus, the point of tangency is (5, 25a).
Now, let's find the slope of the tangent line at that point. Substituting x = 5 into the derivative 2ax, we get the slope as 2a(5) = 10a.
Since the tangent line has the same slope as the derivative of the parabola at the point of tangency, the slope of the tangent line is 10a.
The given equation of the line is 4x + y = b. To find the slope of this line, we can rewrite it in slope-intercept form (y = mx + c), where m is the slope and c is the y-intercept.
Rearranging the equation, we have y = -4x + b, where m (slope) = -4.
For the line to be tangent, the slope of the line, which is -4, must be equal to the slope of the tangent line, which is 10a. So, we can set up the equation:
10a = -4
Solving for a, we get a = -4/10 = -2/5.
Now that we have the value of a, we can substitute it back into the equation of the line to find the value of b.
y = -4x + b
Substituting x = 5 and a = -2/5, we have 25(-2/5) + b = b.
Simplifying further, we get -10 + b = b.
By rearranging the equation, we find b = -10.
Therefore, the values of a and b that make the line 4x + y = b tangent to the parabola y = ax^2 when x = 5 are a = -2/5 and b = -10.
line y=-4a+b slope -4
so if tangent to the curve, y'=-4
y'=-4=2ax=10a
then solve for a, and b.