Let $A B : y = x$ be a chord to the parabola $y^{2} + 4 x = 4$ and $P$ be any variable point on the ellipse $x^{2} + 9 (y - 4)^{2} - 9 = 0 .$ $If Δ_{1} and Δ_{2}$ be the minimum and maximum area of triangle $A B P$ respectively, then

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$Δ_{2} = 2 (4 + \sqrt{10})$

$\frac{Δ_{2}}{Δ_{1}} = \frac{13 + 4 \sqrt{10}}{6}$

$Δ_{2} - Δ_{1} = 8 \sqrt{5}$

$\frac{Δ_{2}}{Δ_{1}} = \frac{13 + 4 \sqrt{10}}{3}$

answer is B, C.

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Detailed Solution

$\begin{aligned} \Rightarrow Δ_{1} = 2 \sqrt{2} (4 - \sqrt{10}) \\ and Δ_{2} = 2 \sqrt{2} (4 + \sqrt{10}) \end{aligned}$

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