Easy calculus question gone hard--- if that makes any sense?

Question

Show that the tangent to any point (a,a^3) on the curve y=x^3 meets the curve again at a point where the slope is four times the slope at (a,a^3)

Answer 1

Well, let's give it a go.
The slope of the curve at (a, a³) is 3a², so the equation
of the tangent line at that point
is y- a³ = 3a²(x-a).
or y = 3a² x -2a³.
Where does this meet the curve again?
When x³ -3a²x + 2a³ = 0.
This looks like a hard equation to solve, but if
you look at it carefully, you see that x = a is a
solution. So x-a is a factor.
Carrying this further, we see that the equation
factors as
(x-a)²(x+2a) = 0.
The double root corresponds to the point of
tangency, so the tangent line to the curve
meets it again when x = -2a.
Thus the point of intersection is (-2a,-8a³).
The slope of y = x³ at that point is 3(4a²),
which is 4 times the slope at (a,a³).
I often use this example with my calculus class
with a = 1 to show that a tangent line to a curve
may meet the curve in another point.

Answer 2

y'=3x^2
the slope at the first point then is 3a^2
so let's find the equation for that line
y=mx+b
a^3=(3a^2)(a)+b = 3a^3 + b
-2a^3 = b
therefore:
y=(3a^2)x - 2a^3

so we want to find when that curve hits y=x^3
so:
x^3 = (3a^2)x - 2a^3
x^3 - (3a^2)x +2a^3 = 0
we know that x=a is a solution since the tangent line hits the curve y=x^3 at x=a. Therefore, we can factor out (x-a) and get:
(x-a)(x^2+ax-2a^2) = 0
(x-a)(x+2a)(x-a) = 0
(x-a)^2 (x+2a) = 0
we don't care about the (x-a) part since we already know it hits there.
so (x+2a)= 0

x=-2a is the point where the tangent line also hits

the slope at x=-2a is :
y'=3x^2
=3(-2a)^2
=12a^2

this slope is exactly 4 times the slope we found for the first point which was 3a^2
4(3a^2) = 12a^2

Answer 3

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Answer 4

y' = 3a^2 at (a, a^3)
y' = 3x^2 = 12a^2 at a point (x,y)
x^2 = 4a^2
Solve for x,
x = ±2a
y = ±8a^3