the tangent line at a point (a,f(a)) is
y= f ' (a) (x-a) +f(a)
for your problem
y=(2-2a)(x-a) + 2a -a^2
For the tangent line to pass through (1,5)
5 = (2-2a)(1-a) + 2a -a^2
solve this for a
See attachment
Here's Fermat's method of finding tangents that pre-dates the calculus.
Any non-vertical line through (1, 5) can be written as y= m(x-1)+ 5. Of course, to be a tangent line, it must touch the parabola at some point: we must have or . A root of that equation would be the x value of the point of tangency. In fact, to be tangent that equation must have a double root. That means we must have for some a. Comparing coefficients, m- 2= -2a and . From the first equation, m= 2- 2a. Putting that into the second equation, . . Use those values of a to find m, the slopes of the two tangent lines, and so the tangent lines.