Equation of the Locus of a Varying Midpoint

Parametric point on the parabola $x^2 = 4ay$ is given by

$(x,y) \equiv (2at, at^2)$

Let the two points $A$ and $B$ be denoted by the parameters $t_1$ and $t_2$ respectively.

Since normal at $A$ meets the parabola again at $B$ , we have a relation between $t_1$ and $t_2$ as

$t_2 = - t_1 - \dfrac{2}{t_1}$

Therefore, points $A$ and $B$ are denoted by

$A \equiv (2at_1 , a{t_1}^2) \\ B \equiv \bigg(-2a\left(t_1 + \dfrac{2}{t_1}\right) , a {\left( t_1 + \dfrac{2}{t_1} \right)}^2 \bigg)$

Let the midpoint of $AB$ be $(h,k)$ . Then

$h = \dfrac{2at_1 - 2a \left( t_1 + \dfrac{2}{t_1} \right)}{2} \\ \\ k = \dfrac{a {\left(t_1 + \dfrac{2}{t_1}\right)}^2 + a{t_1}^2}{2}$

Upon solving

$h = \dfrac{-2a}{t_1} \implies t_1 = \dfrac{-2a}{h}$

Substituting $t_1$ in the value of $k$ we get

$\begin{aligned} & k = \dfrac{a {\left( \dfrac{-2a}{h} - \dfrac{h}{a} \right)}^2 + a {\left( \dfrac{-2a}{h} \right)}^2}{2} \\ \implies & 2k = a \left( \dfrac{4a^2}{h^2} + \dfrac{h^2}{a^2} + 4 \right) + a \left( \dfrac{4a^2}{h^2} \right) \\ \implies & k = \dfrac{8a^4 + h^4 + 4a^2h^2}{2ah^2} \end{aligned}$

Putting the value of $a$ as $a = \dfrac 14$ and changing $(h,k)$ to $(x,y)$ , we get

$\begin{aligned} & y = \dfrac{ 8 \left(\dfrac{1}{256}\right) + x^4 + 4 \left(\dfrac{1}{16}\right)x^2 }{2 \left( \dfrac 14 \right) x^2} \\ \implies & y = \dfrac{32x^4 + 8x^2 + 1}{16x^2} \\ \implies & y = \dfrac{2^5 x^4 + 2^3 x^2 + 1}{2^4 x^2} \end{aligned}$

Thus

$P = 5 \\ Q = 3 \\ R = 4$

Which gives

$P+Q+R= \boxed{12}$