Question

If there are two tangent lines to the curve `y=4x-x^2` that pass through point P(2,5), how do you find the x coordinates of point of tangency?

Answer 1

`(1,3)` and `(3,3)`

Explanation:

The equation of a generic straight passing for `p_0 = {2,5}` is given by
`r->y=5+m(x-2)`
where `m` is the straight declivity.
The intersection points between `r` and the parabola
`p->y=4x-x^2` are found by solving the condition
`5+m(x-2)=4x-x^2` for `x`.
We found
#x = 1/2 (4 - m - sqrt[ m^2-4]), y = 1/2 (10 - m^2 - m sqrt[m^2-4])# and #x = 1/2 (4 - m + sqrt[m^2-4]), y =1/2 (10 - m^2 + m sqrt[m^2-4])#
So each generic straight with declivity `m` intersects the parabola in two points. By conveniently choosing `m` we can make the two points to be coincident.

So, making `m = {-2,2}` we have two straights
`t_1->y=5 - 2 (x-2)` and
`t_2->y=5 + 2 (x-2)`
tangent to the parabola.