We have to find the line passing through (5,2) that is tangential to y = -2x^2 + 4x + 6

Let the required line touch the curve y = -2x^2 + 4x + 6 at (a, b).

The equation of the line is (y - b)/(x - a) = (2 - b)/(5 - a)

The slope of the tangent to the curve -2x^2 + 4x + 6 at (a,b) is -4*a + 4

This gives (2 - b)/(5 - a) = -4*a + 4

=> (2 - b) = 4(1 - a)(5 - a)

Also, as (a, b) lies on the curve y = -2x^2 + 4x + 6, b = -2a^2 + 4a + 6

=> 2 + 2a^2 - 4a - 6 = 4(5 - a - 5a + a^2)

=> 1 + a^2 - 2a - 3 = 2(5 - 6a + a^2)

=> 1 + a^2 - 2a - 3 = 10 - 12a + 2a^2

=> a^2 - 10a + 12 = 0

a1 = [10 + sqrt(100 - 48)]/2

=> a1 = 5 + sqrt 13

a2 = 5 - sqrt 13

b1 = -2(5 + sqrt 13)^2 + 4(5 + sqrt 13) + 6

=> -2(25 + 13 + 10*sqrt 13) + 20 + 4*sqrt 13 + 6

=> -50 - 26 - 20*sqrt 13 + 26 + 4*sqrt 13

=> -50 - 16*sqrt 13

b2 = -2(5 - sqrt 13)^2 + 4(5 - sqrt 13) + 6

=> -2(25 + 13 - 10*sqrt 13) + 20 - 4*sqrt 13 + 6

=> -50 - 26 + 20*sqrt 13 + 26 - 4*sqrt 13

=> -50 + 16*sqrt 13

**There are two lines tangential to the given curve that pass through (5,2): (y - 2)/(x - 5) = (-52/sqrt 13 - 16) and (y - 2)/(x - 5) = (52/sqrt 13 - 16)**