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Locus of Foot of Perpendicular on Chord of Contact | JEE

JEE Maths question with a full step-by-step solution.

Question
Tangents are drawn to y2=4axy^2 = 4ax from a variable point PP moving on x+a=0x + a = 0. Then the locus of the foot of the perpendicular drawn from PP on the chord of contact of PP is:
Ay=0y = 0
B(x−a)2+y2=a2(x - a)^2 + y^2 = a^2
C(x−a)2+y2=0(x - a)^2 + y^2 = 0correct
Dy(x−a)=0y(x - a) = 0
Solution
Step 1: For P(−a,p)P(-a, p) on the directrix, the chord of contact is py=2a(x−a)py = 2a(x - a), i.e. 2ax−py−2a2=02ax - py - 2a^2 = 0, with normal direction (2a,−p)(2a, -p). Step 2: The chord value at PP is −(4a2+p2)-(4a^2 + p^2), exactly −1-1 times ∣(2a,−p)∣2|(2a, -p)|^2, so the foot of perpendicular is
N=P+(2a,−p)=(a,0)N = P + (2a, -p) = (a, 0)
 for every pp. The locus is the single point (x−a)2+y2=0(x - a)^2 + y^2 = 0. Correct answer: (3)
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