The tangent at $P$ to $y = x^2 - x^3$ meets the curve again at $Q$. Show that locus of midpoint of $PQ$ is $y=1-9x+28x^2-28x^3$

Question

The tangent at a variable point $P$ of the curve $y = x^2 - x^3$ meets it again at $Q$. Show that the locus of the middle point of $PQ$ is $$y = 1 - 9x + 28x^2 - 28x^3$$

My approach

$$y^\prime=2x-3x^2$$

Equation of tangent: $$2\ x_1 -3\ x_1^2=\frac{y-\ x_1^2+\ x_2^3}{x-\ x_1}$$

From the equation of tangent: $$(2\ x_1 -3\ x_1^2)(x-\ x_1)=y-\ x_1^2+\ x_2^3$$

Substituting the value of $y$ in the conic section we can find the value of $x$ but still it is getting complicated.

Answer 1

Given $f(x) = x^2-x^3$ the tangent line at $x_0$ is

$$ y = f(x_0) + f'(x_0)(x-x_0)\to y = \left(2 x_0-3 x_0^2\right) (x-x_0)-x_0^3+x_0^2 $$

Solving now for the intersection points we have

$$ y = x^2-x^3\\ y = \left(2 x_0-3 x_0^2\right) (x-x_0)-x_0^3+x_0^2 $$

with the feasible solution $x = 1-2x_0$ so the sought locus is given by

$$ (x, y) = \frac 12((x_0, f(x_0))+(1-2x_0,f(1-2x_0))) = \frac 12(1-x_0, 7(x_0^3-x_0^2)+2x_0) $$

Attached a plot showing $f(x)$ in blue, $\frac 12(1-x_0, 7(x_0^3-x_0^2)+2x_0)$ in dashed red and the tangent line.

NOTE

Eliminating $x_0$ in the equations

$$ (x,y) = \frac 12(1-x_0, 7(x_0^3-x_0^2)+2x_0) $$

we obtain

$$ y = 1-9x+28x^2-28 x^3 $$

Answer 2

Let $P(h,h^2-h^3),Q(k,k^2-k^3), h\ne k$

Now the gradient of $PQ$ is $$=\dfrac{k^2-k^3-(h^2-h^3)}{k-h}=k+h-(k^2+kh+h^2)$$

The gradient of the tangent at $P$ is $$=2h-3h^2$$

$\implies2h-3h^2=k+h-(k^2+kh+h^2)$

$\iff2h^2-(k+1)h+k-k^2=0$

$$\implies h=\dfrac{k+1\pm\sqrt{(k+1)^2+8(k^2-k)}}4=\dfrac{k+1\pm(3k-1)}4=k,\dfrac{1-k}2$$

As $h\ne k, h=\dfrac{1-k}2$

If $R(p,q)$ is the midpoint, $$2p=h+k=h+(1-2h)\iff h=?$$

$2q=h^2+k^2-(h^3+k^3)=h^2+(1-2h)^2-h^3-(1-2h)^3$

Replace the value of $h$ in terms of $p$