If two distinct chords of a parabola $y^{2} = 4ax$ passing through the point $(a,2a)$ are bisected b

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If two distinct chords of a parabola $y^{2} = 4ax$ passing through the point $(a,2a)$ are bisected by line $x + y = 1$, then the length of the latus rectum can not be :

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Sol. [Key Concept: A chord with midpoint $(h,k)$ has equation $T = S_{1}$; two distinct such chords through $(a,2a)$ bisected on $x + y = 1$ exist only when a quadratic in the midpoint has two real roots — that bounds the latus rectum.]Put $(2h - a,2 - 2h - 2a)$ in the equation of parabola$${\Rightarrow 4(1 - h - a)^{2} = 4a(2h - a) }{\Rightarrow h^{2} - 2h + \left( 2a^{2} - 2a + 1 \right) = 0 - h_{1} }$$For two distinct real roots$$D > 0 \Rightarrow 4 - 4\left( 2a^{2} - 2a + 1 \right) > 0$$$${\Rightarrow a(a - 1) < 0 }{\Rightarrow a \in (0,1) }$$⇒ Length of Latus rectum $\in (0,4)$

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