Three circles, each of radius 1, are packed inside an ellipse x2a2+y2b2=1(a>b>0), (2θ=∠ABC), as shown in the fig. Two of these circles, touch the ellipse at two points each, and the middle circle touches it at the end of minor axis. A, B and C are centres of these circles. If the area of ellipse is minimum, then which of the following options is/are correct?

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Three circles, each of radius 1, are packed inside an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b > 0), (2 θ = ∠ A B C),$
as shown in the fig. Two of these circles, touch the ellipse at two points each, and the middle circle touches it at the end of minor axis. A, B and C are centres of these circles.
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If the area of ellipse is minimum, then which of the following options is/are correct?

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3b = 4

$\sqrt{6} e = \sqrt{5}$

$6 \tan θ = \sqrt{35}$

$a = 4 \sqrt{3}$

answer is B, C.

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Detailed Solution

$E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$a r e a, Δ = π a b O B J : Δ = f (θ) ∴ b - 2 \cos θ = 1 \Rightarrow b = 1 + 2 \cos θ$

Also normal to ellipse at H passes through C

$L e t H (p, q) \Rightarrow N @ H \equiv \frac{a^{2} x}{p} - \frac{b^{2} y}{q} = a^{2} - b^{2} = a^{2} e^{2} \Rightarrow c (p e^{2}, 0)$

$p e^{2} = 2 \sin θ a n d b - 1 = 2 \cos θ ∴ {(p e^{2})}^{2} + {(b - 1)}^{2} = 4$

$H C = 1 ∴ {(p - 2 \sin θ)}^{2} + q^{2} = 1$

$H C = 1 ∴ {(p - 2 \sin θ)}^{2} + q^{2} = 1 ∴ {(p (1 - e^{2}))}^{2} + q^{2} = 1$

$a l s o H l i e s o n e l l i p s e \Rightarrow \frac{p^{2}}{a^{2}} + \frac{q^{2}}{b^{2}} = 1 \Rightarrow p^{2} (1 - e^{2}) + q^{2} = b^{2}$

$p^{2} e^{2} (1 - e^{2}) = b^{2} - 1 a n d p^{2} e^{4} = 4 - {(b - 1)}^{2}$

$\Rightarrow \frac{1 - e^{2}}{e^{2}} = \frac{b^{2} - 1}{3 - b^{2} + 2 b} \Rightarrow 1 - e^{2} = \frac{b^{2} - 1}{2 + 2 b} = \frac{b^{2}}{a^{2}}$

$\Rightarrow a^{2} = \frac{2 b^{2}}{b - 1} \Rightarrow Δ = π a b = \frac{π b^{2} \sqrt{2}}{\sqrt{b - 1}} = \frac{π \sqrt{2}}{\sqrt{b^{- 3} - b^{- 4}}}$