The tangent to the curve C: y=4x^2 at the point A is parallel to the straight line L: 2x-y+24=0. Find the equation of the tangent to C at A.

To find the equation of the tangent to the curve C at point A, we first need to find the derivative of the function y = 4x^2. The derivative of this function is y' = 8x. This derivative gives us the slope of the tangent line at any point on the curve.

The line L: 2x - y + 24 = 0 can be rewritten in slope-intercept form as y = 2x + 24. The slope of this line is 2.

Since the tangent to the curve C at point A is parallel to line L, it must have the same slope. Therefore, we set the derivative equal to 2:

8x = 2 x = 2 / 8 x = 0.25

Now that we have the x-coordinate of point A, we can find the y-coordinate by substituting x = 0.25 into the equation of the curve:

y = 4(0.25)^2 y = 4 * 0.0625 y = 0.25

So, point A is (0.25, 0.25).

The equation of the tangent line is given by the point-slope form of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Substituting m = 2, x1 = 0.25, and y1 = 0.25, we get:

y - 0.25 = 2(x - 0.25) y = 2x - 0.5 + 0.25 y = 2x - 0.25

So, the equation of the tangent to the curve C at point A is y = 2x - 0.25.