# Thread: equation of tangent lines

1. ## equation of tangent lines

Find the equation of 2 tangent lines to the graph of f(x)=2x-x^2 that passes through the point (1,5)

2. the tangent line at a point (a,f(a)) is

y= f ' (a) (x-a) +f(a)

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

3. 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 $m(x- 1)+ 5= 2x- x^2$ or $x^2+ (m-2)x+ 5-m=0$. 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 $x^2+ (m-2)x+ 5-m= (x- a)^2= x^2- 2ax+ a^2$ for some a. Comparing coefficients, m- 2= -2a and $5- m= a^2$. From the first equation, m= 2- 2a. Putting that into the second equation, $5- (2- 2a)= 3+ 2a= a^2$. $a^2- 2a- 3= (a- 3)(a+ 1)= 0$. Use those values of a to find m, the slopes of the two tangent lines, and so the tangent lines.

4. I would have been a serf in those days