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Locus of Centre of Focal Chord Circle on y²=4ax | JEE

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

Question
The locus of the centre of the circle described on any focal chord of the parabola y2=4axy^2 = 4ax as diameter is:
Ax2=2a(y−a)x^2 = 2a(y - a)
Bx2=−2a(y−a)x^2 = -2a(y - a)
Cy2=2a(x−a)y^2 = 2a(x - a)correct
Dy2=−2a(x−a)y^2 = -2a(x - a)
Solution
Step 1: The endpoints (at12,2at1),(at22,2at2)(at_1^2, 2at_1),(at_2^2, 2at_2) of a focal chord have t1t2=−1t_1 t_2 = -1, and the centre is their midpoint (h,k)(h, k):
k=a(t1+t2),h=a2(t12+t22).k = a(t_1 + t_2),\qquad h = \frac{a}{2}(t_1^2 + t_2^2).
 Step 2: Using t1+t2=kat_1 + t_2 = \dfrac{k}{a} and t1t2=−1t_1 t_2 = -1,
t12+t22=k2a2+2⇒h=k22a+a⇒y2=2a(x−a).t_1^2 + t_2^2 = \frac{k^2}{a^2} + 2 \Rightarrow h = \frac{k^2}{2a} + a \Rightarrow y^2 = 2a(x - a).
 Correct answer: (3)
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