We have,
y=4x2+1
L:y=x
Let the foot of perpendicular from P to line y=x is Q.
Let P≡(x,y), Q≡(c,c) and R≡(h,k) where, R is the mid-point of PQ.

Clearly,
PQ⊥L
⇒h−ck−c=−1
⇒c=(2h+k)
And,
R≡(2x+c,2y+c)
⇒R≡(2x+4h+4k,2y+4h+4k)
Hence,
h=2x+4h+4k⇒x=23h−2k
k=2y+4h+4k⇒y=23k−2h
Now,
y=4x2+1
⇒(23k−h)=4(23h−k)2+1
⇒(3k−h)=2(3h−k)2+2
Required locus is
2(3x−y)2+(x−3y)+2=0