To find the points of intersection between the line 4x - 7y + 10 = 0 and the parabola y^2 = 4x, we need to solve these two equations simultaneously.

1. Solving the equations:
To solve the equations, we can substitute the value of x from the first equation into the second equation.

From the first equation, we have:
4x - 7y + 10 = 0
4x = 7y - 10
x = (7y - 10)/4

Substituting this value of x into the second equation:
y^2 = 4(7y - 10)/4
y^2 = 7y - 10
y^2 - 7y + 10 = 0

Now we can factorize the quadratic equation:
(y - 2)(y - 5) = 0

So, the values of y are y = 2 and y = 5.

2. Finding the corresponding x-coordinates:
To find the corresponding x-coordinates, we substitute the values of y into the equation x = (7y - 10)/4.

For y = 2:
x = (7(2) - 10)/4
x = (14 - 10)/4
x = 4/4
x = 1

For y = 5:
x = (7(5) - 10)/4
x = (35 - 10)/4
x = 25/4

So, the points of intersection are A(1, 2) and B(25/4, 5).

3. Finding the tangents at points A and B:
To find the tangents at points A and B, we need to find the slope of the tangent at each point. The slope of the tangent is equal to the derivative of the equation of the parabola at that point.

The derivative of the equation y^2 = 4x is:
dy/dx = 2√x

At point A(1, 2):
dy/dx = 2√1 = 2

At point B(25/4, 5):
dy/dx = 2√(25/4) = 2√(25/4) = 2√(25)/√(4) = 2(5)/2 = 5

4. Finding the equations of the tangents:
Using the point-slope form of a line, we can find the equations of the tangents at points A and B.

For point A(1, 2):
y - 2 = 2(x - 1)
y - 2 = 2x - 2
y = 2x

For point B(25/4, 5):
y - 5 = 5(x - 25/4)
y - 5 = 5x - 125/4
y = 5x - 125/4 + 20/4
y = 5x - 105/4

5. Finding the point of intersection of the tangents:
To find the point of intersection of the tangents, we can solve the two tangent equations simultaneously.

By equating the two equations:
2x = 5x - 105/4
3x =