If $R (2, 6), S (12, K)$ are the foci of an ellipse which touches both the coordinate axes then

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Length of its latusrectum is $\frac{24 \sqrt{2}}{5}$

Its eccentricity $\frac{\sqrt{13}}{5}$

Length of its double ordinate bisecting $\bar{R S}$ is $4 \sqrt{6}$

Length of its major axis is $5 \sqrt{2}$

answer is A, B, D.

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Detailed Solution

$x = 0, y = 0$ are tangents
$∴ b^{2} = 2.12 = 6. K \Rightarrow b^{2} = 24, K = 4$ ( product of perpendicular distances from foci to tangents)
$2 a e =$ distance between foci $= \sqrt{10^{2} + 2^{2}} = \sqrt{104}$
$4 (a^{2} - b^{2}) = 104 \Rightarrow a^{2} = 50$
$∴ e = \sqrt{\frac{50 - 24}{50}} = \frac{\sqrt{13}}{5}$
LLR $= \frac{2 b^{2}}{a} = \frac{2.24}{5 \sqrt{2}} = \frac{24 \sqrt{2}}{5}$
Length of major axis $= 2 a = 10 \sqrt{2}$
Double ordinate is minor axis
Length $= 2 b = 4 \sqrt{6}$