Let R(a,b) be mid-point of line joining point P and Q where PQ is perpendicular to line y=x. Let coordinates of P be P(x,y),Q(q,q) and R(a,b) then, a=
x+q
2
and b=
y+q
2
Now, slope of line y=x is m1=1 Slope of line PO be
b−q
a−q
=m2 (say) ∵ Line y=x and PQ are perpendicular to each other, m1⋅m2=−1 ⇒
b−q
a−q
=−1⇒b−q=q−a ⇒q=
b+a
2
∴a=
x+q
2
=
x+(
b+a
2
)
2
=
2x+b+a
4
⇒x=
4a−b−a
2
=
3a−b
2
and b=
y+q
2
=
y+(
b+a
2
)
2
=
2y+b+a
4
⇒y=
3b−a
2
Put (x,y) in equation of parabola as P(x,y) is variable point on parabola
3b−a
2
=4(
3a−b
2
)2+1
(3b−a)
2
=(3a−b)2+1 ⇒(3b−a)=2(3a−b)2+2 Replace (a,b) as (x,y) ⇒(3y−x)=2(3x−y)2+2 or 2(3x−y)2+(x−3y)+2=0