How to find the equation of a tangent line?

Question

There are two lines that pass through the point (3,0) and are tangent to the parabola f(x) = 9 x^2

One of those two lines is the x-axis. Find the equation for the other line. (Suggestion: First find the value of a shown on the graph).

I'm quite confused with this. Any suggestions?

Answer 1

f(x) = 9 x^2
f'(x) = 18 x
18x = (9x^2 - 0)/(x - 3)
18x^2 - 54x = 9x^2
9x^2 - 54x = 0
x(x - 6) = 0
x = 6
18x = 108

y = 108(x - 3)
y = 108x - 324

Answer 2

y = 9x^2
dy/dx = 18x

Coordinates of any point on the curve will be (x, 9x^2)

Slope of line (rise/run) joining (x,9x^2) and (3,0) is (9x^2)/(x-3)

(9x^2)/(x-3) = 18x
9x^2 = 18x^2 - 54x
9x^2 - 54x = 0
x^2 - 6x = 0
x(x-6) = 0
x = {6, 0}
y = 6x^2 = {216, 0}

The coordinates on the graph are (6,216) and (0,0)
The line passing through (0,0) and (3,0) is the x-axis

The line passing through (6,216) and (3,0):
Slope = 216/(9) = 24

y = mx + c
m = 24
y = 0
x = 3
0 = 72 + c
c = -72

The equation of the required line is
y = 24x - 72
24x - y = 72