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Parabola: Tangents Drawn Variable Point Parabola 4ax Points They

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

Question
Two tangents are drawn from a variable point PP to the parabola y2=4axy^{2} = 4ax, and the points where they meet the co-ordinate axes are concyclic. Given that the locus of PP is x=λ1ax = \lambda_1 a, find λ1\lambda_1.
Solution
Answer: 1
Question attachment Step 1: The tangent at (at2,2at)\left(at^{2},2at\right) is
ty=x+at2ty = x + at^{2}
 Let the two tangents from PP have parameters t1t_1 and t2t_2. Step 2: The first tangent cuts the axes at
x=0⇒y=at1⇒T(0, at1);y=0⇒x=−at12⇒R(−at12, 0)x = 0 \Rightarrow y = at_1 \Rightarrow T\left(0,\,at_1\right); \qquad y = 0 \Rightarrow x = -at_1^{2} \Rightarrow R\left(-at_1^{2},\,0\right)
 Similarly the second tangent gives Q(0, at2)Q\left(0,\,at_2\right) and S(−at22, 0)S\left(-at_2^{2},\,0\right). Step 3: T,QT,Q lie on the yy-axis and R,SR,S on the xx-axis, and the two axes meet at the origin OO. Four such points are concyclic exactly when OO has the same power along both lines, the segments being taken with sign:
OT⋅OQ=OR⋅OSOT\cdot OQ = OR\cdot OS
 Step 4: Substituting the signed lengths,
(at1)(at2)=(−at12)(−at22) ⟹ a2t1t2=a2(t1t2)2\left(at_1\right)\left(at_2\right) = \left(-at_1^{2}\right)\left(-at_2^{2}\right) \ \Longrightarrow\ a^{2}t_1t_2 = a^{2}\left(t_1t_2\right)^{2}
 Neither tangent passes through the vertex, so t1t2≠0t_1t_2 \ne 0, and dividing by a2t1t2a^{2}t_1t_2,
t1t2=1t_1t_2 = 1
 Step 5: The tangents at t1t_1 and t2t_2 meet at
P(at1t2, a(t1+t2))P\left(at_1t_2,\ a(t_1+t_2)\right)
 so the abscissa of PP is x=at1t2=ax = at_1t_2 = a. Step 6: The locus is the line x=a=1⋅ax = a = 1\cdot a, so λ1=1\lambda_1 = 1. Answer: 11.
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