Tangent at Parameter $$t$$

For y²=4x (so a=1), general point is (t², 2t).
P = (t₁², 2t₁), Q = (t₂², 2t₂)

Condition: PQ subtends 90° at O(0,0)
Slope of OP = 2t₁/t₁² = 2/t₁. Slope of OQ = 2/t₂. For perpendicularity:
(2/t₁) × (2/t₂) = −1
4/(t₁t₂) = −1
t₁t₂ = −4

Tangent to y²=4x at (t², 2t) is: ty = x + t²
Tangent at P: t₁y = x + t₁² ...(i)
Tangent at Q: t₂y = x + t₂² ...(ii)

Subtract (ii) from (i):
(t₁ − t₂)y = t₁² − t₂² = (t₁−t₂)(t₁+t₂)
Y = t₁ + t₂ [since t₁ ≠ t₂]

Substitute Y into (i):
t₁(t₁+t₂) = X + t₁²
t₁² + t₁t₂ = X + t₁²
X = t₁t₂ = −4

T = (−4, t₁+t₂)

The x-coordinate of T is always −4, for any t₁, t₂ satisfying t₁t₂=−4. The y-coordinate t₁+t₂ takes all real values.
Locus: X = −4
⟹ x + 4 = 0