1. Take a variable point on the parabola

Let $P(t,\,4t^2+1)$ be any point on the parabola $y=4x^2+1.$

2. Find the foot of perpendicular from $P$ to the line $y=x$

The line $y=x$ has the property that the reflection/projection formulas are simple.

If a point is $(a,b)$, then its foot of perpendicular on $y=x$ is $\left(\frac{a+b}{2},\frac{a+b}{2}\right).$

So for $P(t,4t^2+1),$ the foot of perpendicular $F$ on $y=x$ is $F\left(\frac{t+4t^2+1}{2},\frac{t+4t^2+1}{2}\right).$

3. Find the midpoint of $P$ and $F$

Let the midpoint be $M(x,y)$. Then $x=\frac{1}{2}\left[t+\frac{t+4t^2+1}{2}\right],$ $y=\frac{1}{2}\left[(4t^2+1)+\frac{t+4t^2+1}{2}\right].$

Simplify: $x=\frac{2t+t+4t^2+1}{4}=\frac{4t^2+3t+1}{4},$ $y=\frac{8t^2+2+t+4t^2+1}{4}=\frac{12t^2+t+3}{4}.$

Thus, $x=\frac{4t^2+3t+1}{4},\qquad y=\frac{12t^2+t+3}{4}.$

4. Eliminate the parameter $t$

Compute the combinations appearing in the options.

(i) Find $3x-y$

$3x-y=3\cdot \frac{4t^2+3t+1}{4}-\frac{12t^2+t+3}{4}$ $=\frac{12t^2+9t+3-12t^2-t-3}{4}$ $=\frac{8t}{4}=2t.$

So, $t=\frac{3x-y}{2}.$

(ii) Find $x-3y$

$x-3y=\frac{4t^2+3t+1}{4}-3\cdot \frac{12t^2+t+3}{4}$ $=\frac{4t^2+3t+1-36t^2-3t-9}{4}$ $=\frac{-32t^2-8}{4}=-8t^2-2.$

Now substitute $t=\frac{3x-y}{2}.$ Then $t^2=\frac{(3x-y)^2}{4}.$

Hence, $x-3y=-8\left(\frac{(3x-y)^2}{4}\right)-2$ $x-3y=-2(3x-y)^2-2.$

Rearranging, $2(3x-y)^2+(x-3y)+2=0.$

5. Match with the options

This is exactly Option B: $2{(3x-y)^2} + (x-3y)+2=0.$

6. Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

So the derived answer agrees with the stored correct answer.