Find equation of a line (that runs through 3,0) that is a tangent to a parabola (which has equation y=6x^2)?

Question

I have to different answers....should I believe the top contributor?

Answer 1

y = 6x²

take first derivative to get slope of tangent

y' = 12x

slope m of line passing through (3,0) and (x,y)

m = y/(x-3)

= 6x²/(x-3)

12x = 6x²/(x-3)

12x(x-3) = 6x²

2x(x-3) = x²

2x²-6x = x²

x² - 6x = 0

x(x-6) = 0

x = 0, 6

solutions: (0,0) and (6,216)

The line passing through (3,0) and (0,0) has slope (0-0)/(0-3) = 0 and y-intercept 0. Equation of line:

y = 0

The line passing through (3,0) and (6,216) has slope (216-0)/(6-3) = 72 and y-intercept 216-72(6) = -216. Equation of line:

y = 72x - 216

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Answer 2

first find the graient of the parabola at the point (3,0) as this will be the gradient of the line. (since it is a tangent).

dy/dx = 12x (the gradient function using calculus)

when x=3 (from point 3,0)

dy/dx = 12 x 3 = 36

you now know the gradient of the line and a point it passes through. Now use this formula for a straight line:

y - y1 = m(x - x1)

y and x are the variables. y1 is the value of y in the point known and similiarly for x1 which is the value of x in the point known. m is the gradient.

sub into the formula and you get:

y - 0 = 36(x - 3)

y = 36x - 108

Answer 3

Let the tangent touch the parabola at (x1, y1)

y = 6x²

dy/dx = 12x

At (x1, y1):

dy/dx = 12x1

y1 = 6(x1)²

Slope of the line joining (3, 0) with (x1, y1) is:

y1/(x1 - 3)

y1/(x1 - 3) = 12x1

6(x1)² / (x1 - 3) = 12x1

x1/(x1 - 3) = 2

2x1 - 6 = x1

x1 = 6

y1 = 216

The slope of the line is 72

y = mx + c

m = 72

x = 3

y = 0

c = -216

y = 72x - 216

72x - y = 216