Problem 766

When the line $y=-15$ intersects the parabola with equation $y=-x^2+2x$, the $x$-coordinates of the two points of intersection satisfy the equation $-15=-x^2+2x$.

Solving this equation, we get $x^2-2x-15=0$ or $(x+3)(x-5)=0$, and so $x=-3$ or $x=5$.
Since both points of intersection lie on the line $y=-15$, then the coordinates of the two points of intersection are $(-3,-15)$ and $(5,-15)$.
The point with $x$-coordinate 4, on the parabola with equation $y=-x^2-3x$, has $y$-coordinate $-4^2-3(4)=-28$.

Therefore, the line intersects the parabola at the point $(4,-28)$.

The line passes through the point $(0,8)$, and so the line has slope $\dfrac{-28-8}{4-0}=\dfrac{-36}{4}=-9$.

The line has slope $-9$ and $y$-intercept 8, and so the equation of the line is $y=-9x+8$.

When the line $y=-9x+8$ intersects the parabola with equation $y=-x^2-3x$, the $x$-coordinates of the two points of intersection satisfy the equation $-9x+8=-x^2-3x$.

Solving this equation, we get $x^2-6x+8=0$ or $(x-2)(x-4)=0$, and so $x=2$ or $x=4$.
Therefore, the line intersects the parabola at $x=4$ and at $x=2$, and so $a=2$.
The point with $x$-coordinate $p$, on the parabola with equation $y=-x^2+kx$, has $y$-coordinate $-p^2+kp$.

Therefore, the line intersects the parabola at the point $(p,-p^2+kp)$.

Similarly, the line also intersects the parabola at the point $(q,-q^2+kq)$.
The slope of the line passing through the points $(p,-p^2+kp)$ and $(q,-q^2+kq)$ is $\dfrac{(-p^2+kp)-(-q^2+kq)}{p-q}$, where $p\neq q$ and so $p-q\neq0$.

Simplifying this slope, we get $$\begin{align*} \dfrac{(-p^2+kp)-(-q^2+kq)}{p-q}&=\dfrac{q^2-p^2+kp-kq}{p-q}\\ &=\dfrac{(q-p)(q+p)+k(p-q)}{p-q}\\ &=\dfrac{(q-p)(q+p)}{p-q}+\dfrac{k(p-q)}{p-q}\\ &=\dfrac{-(p-q)(q+p)}{p-q}+\dfrac{k(p-q)}{p-q}\\ &=-(q+p)+k\\ &=k-q-p\end{align*}$$

The line has slope $k-q-p$ and passes through the point $(p,-p^2+kp)$.

Therefore, the equation of the line is $y-(-p^2+kp)=(k-q-p)(x-p)$.

(The equation of a line having slope $m$ and passing through the point $(x_1,y_1)$ is $y-y_1=m(x-x_1)$. This is called the point-slope form of a line.)
Finally, we determine the $y$-intercept of the line by substituting $x=0$ into the equation of the line $y-(-p^2+kp)=(k-q-p)(x-p)$ and solving for $y$.

$$\begin{align*} y-(-p^2+kp)&=(k-q-p)(x-p)\\ y-(-p^2+kp)&=(k-q-p)(0-p)\\ y+p^2-kp&=-kp+pq+p^2\\ y&=-p^2+kp-kp+pq+p^2\\ y&=pq\end{align*}$$

The $y$-intercept of the line that intersects the parabola with equation $y=-x^2+kx$ at $x=p$ and at $x=q$ with $p\neq q$, is $pq$.
When the curve $x=\dfrac{1}{k^3}y^2+\dfrac{1}{k}y$ intersects the parabola with equation $y=-x^2+kx$,
the $x$-coordinates of the two points of intersection ($(0,0)$ and $T$) satisfy the equation $x=\dfrac{1}{k^3}(-x^2+kx)^2+\dfrac{1}{k}(-x^2+kx)$, where $k\neq0$.

Simplifying this equation, we get $$\begin{align*} x&=\dfrac{1}{k^3}(-x^2+kx)^2+\dfrac{1}{k}(-x^2+kx)\\ k^3x&=(-x^2+kx)^2+k^2(-x^2+kx)\\ k^3x&=x^4-2kx^3+k^2x^2-k^2x^2+k^3x\\ 0&=x^4-2kx^3\\ 0&=x^3(x-2k)\end{align*}$$

Since $x^3(x-2k)=0$, then the $x$-coordinates of the points of intersection of the curve and the parabola are $x=0$ and $x=2k$.

Therefore, the $x$-coordinate of point $T$ is $x=2k$, and the $y$-coordinate of $T$ is $-(2k)^2+k(2k)=-4k^2+2k^2=-2k^2$.

Since the $y$-coordinate of point $T$ does not contain a linear term in the variable $k$ and does not contain a constant term, then the equation of the parabola on which all such points $T$ lie, contains a quadratic term only.

That is, all points $T(2k,-2k^2)$ lie on a parabola with an equation of the form $y=ax^2$.

Substituting, we get $-2k^2=a(2k)^2$ or $-2k^2=4ak^2$ or $-2=4a$ (since $k\neq0$), and so $a=-\dfrac12$.

(We may verify that $x=2k$ and $y=-2k^2$ satisfies $y=ax^2+bx+c$ only if $a=-\dfrac12$ and $b=c=0$.)
Therefore, the equation of the required parabola is $y=-\dfrac{1}{2}x^2$.