By drawing a diagram, we can more clearly identify the curve $D$.

The initial curve has equation $y=x^2$, so the gradient of PQ (normal to the curve) is given by $$m_{PQ} = \left. \frac{-2}{\frac{dy}{dx}}\right|_{x=-t} = \frac{1}{2t}$$ The equation of the line $PQ$ can therefore be expressed as $$y - t^2 = \frac{1}{2t}(x+t)$$ To find the point Q, we need to solve the simultaneous equation $$y - t^2 = \frac{1}{2t}(x+t),\;y=x^2$$ We get $x^2 - \frac{x}{2t} - t^2 - \frac{1}{2} = 0$ which is a quadratic in $x$. Since we know that x = -t is already a solution, it is trivial to see that the other solution is $x = t + \frac{1}{2t}$ Consequently the two points of intersection are $$ P(-t,t^2), \; Q\left(t+ \frac{1}{2t},\left(t+ \frac{1}{2t}\right)^2\right) $$ The midpoint is therefore $$ M = \left(\frac{1}{4t}, \frac{t^2 + \left(t+\frac{1}{2t}\right)^2}{2}\right) $$ Let $x = \frac{1}{4t}$, then we get the equation of the curve is $$ y = \frac{\left(\right)^2 + \left(\frac{1}{4x} + 2x\right)^2}{2} $$ Simplifying this gives $$ y = \frac{1}{32x^2} + \frac{1}{32x} + \frac{1}{2} + 2x^2 $$