find the equation of the line.

Question

Let f(x)=x^2 and g(x)=(x-74)^2+27. There is one line with positive slope that is tangent to both of the parabolas y=f(x) and y=g(x) simultaneously.

 

 

Answer 1

Let (c,c²) be the point of tangency on the graph of f(x) = x^2.

 

The graph of g(x) = (x-74)² + 27 can be obtained by translating the graph of f(x) = x² 74 units to the right and 27 units upward.

 

So, the point (c+74, c²+27) is the corresponding point of tangency on the graph of y = g(x),

 

Slope of line = [c²+27 - c²] / [c+74 - c] = 27/74

 

But, the slope of the line can also be expressed as f'(c) = 2c.

 

So, 2c = 27/74.   c = 27/148

 

Point on line:  (c, c²) = (27/148, 729/21904)

 

Equation of line:  y - (729/21904) = (27/74)(x - 27/148)