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Match the items of Column – I with Column – II
Column – I Column – II
I If the coordinates of a point are $(4 \tan θ, 3 \sec θ)$ where $θ$ is parameter then the points lies on a conic whose eccentricity is P $\sqrt{3}$
II Tangent at any point P on an ellipse whose foci are $F_{1}$ , $F_{2}$ meets the Auxiliary circle of the ellipse at $B_{1}$ and $B_{2}$ . If $F_{1} P + F_{2} P = 10$ and $(F_{1} B_{1}) (F_{2} B_{2}) = 16$ then eccentricity of ellipse Q $\frac{4}{5}$
III If AB is double ordinate of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{9} = 1$ such that triangle OAB is an equilateral triangle of side ‘2’ then eccentricity of hyperbola is (where O is centre of Hyperbola) R $\frac{5}{3}$
IV If the foci of ellipse $\frac{x^{2}}{K^{2} a^{2}} + \frac{y^{2}}{a^{2}} = 1$ and hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}} = 1$ coincide then K can be S $\frac{3}{5}$
T $\sqrt{\frac{13}{3}}$

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answer is A.

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Detailed Solution

I) $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$
$\frac{x^{2}}{16} - \frac{y^{2}}{9} = - 1$
$e = \sqrt{\frac{16 + 9}{16}} = \frac{5}{4}$
$\frac{1}{e^{2}} + \frac{1}{e'^{2}} = 1$
$\frac{1}{e'^{2}} = 1 - \frac{16}{25} = \frac{9}{25}$
$e^{1} = \frac{5}{3}$
II) 2a=10 $\Rightarrow$ a=5
$b^{2} = 16$
b=4
$e = \sqrt{\frac{25 - 16}{25}}$
III) $(a \sec θ, 3 \tan θ)$
$6 \tan θ = 2$
$\tan θ = \frac{1}{3}$
$\frac{1}{\sqrt{3}} = \frac{3 \tan θ}{a \sec θ}$
   $a^{2} \sec^{2} θ = 27 \tan^{2} θ$
$a^{2} (1 + \frac{1}{9}) = 27. \frac{1}{9}$
   $a^{2} (10) = 27$
IV) $k^{2} a^{2} - a^{2} = a^{2} + a^{2}$
$k^{2} - 1 = 2 \Rightarrow k^{2} = 3 \Rightarrow \sqrt{3}$

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