Two perpendicular chords $A B$ and $A C$ are drawn through the vertex $A$ of the parabola $y^{2} = 4 a x$ . The locus of circumcentre of the $△ A B C$ is $y^{2} = 2 a x - λ a^{2}$ . Then, the value of $λ =$

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Solution

Let B and C be $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ and the $(α, β)$ be the circumcentre of $Δ A B C$ , then
$α = \frac{a t_{1}^{2} + a t_{2}^{2}}{2}, β = \frac{2 a t_{1} + 2 a t_{2}}{2}$
( $∵$ ABC is a right angle triangle)
Also, $A B ⊥ A C$
$\frac{2}{t_{1}} \times \frac{2}{t_{2}} = - 1 \Rightarrow t_{1} t_{2} = - 4$

Elimination of $t_{1}$ and $t_{2}$ yields
$y^{2} = 2 a x - 8 a^{2} ∴ λ = 8$

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