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Match the items of Column – I with Column – II
Column I Column II
I A variable tangents to $\frac{x^{2}}{4} - y^{2} = 1$ cuts $\frac{x^{2}}{4} + y^{2} = 1$ in points P &Q .Locus of mid-point of PQ is ${(x^{2} + 4 y^{2})}^{2} = K {(x^{2} - 4 y^{2})}^{2}$ then K= P 1
II From a point (0,2) a pair of tangents are drawn to $x^{2} - y^{2} = 1$ whose magnitude of the product of slopes Q 4
III A circle of radius 2 unit is drawn through the points in which $\frac{x^{2}}{3} - \frac{y^{2}}{2} = 1$ & $\frac{x^{2}}{a^{2}} + y^{2} = 1$ intersect. Then $a^{2} =$ R 5
IV If a variable line has intercepts on the X &Y axes $e & e'$ respectively; where $e & e'$ are the eccentricities of a hyperbola and its conjugate hyperbola, then this line always touches the circle $x^{2} + y^{2} = r^{2}$ , where r= S 6

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$I - R; I I - Q; I I I - S; I V - P$

$I - Q; I I - R; I I I - P; I V - S$

$I - R; I I - Q; I I I - P; I V - S$

$I - Q; I I - R; I I I - S; I V - P$

answer is B.

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Detailed Solution

(i) Let the locus be . So $\frac{x h}{4} + y k = \frac{h^{2}}{4} + k^{2}$
$\Rightarrow y = \frac{- x h}{4 k} + \frac{h^{2}}{4 k} + k$
Hence, ${(\frac{h^{2}}{4 k} + k)}^{2} = \frac{4 h^{2}}{16 k^{2}} - 1 \Rightarrow (h^{2} + 4 k^{2}) = 4 (h^{2} - 4 k^{2})$
(ii) Let $y = m x + 2$ be a tangent. So, $4 = m^{2} - 1 \Rightarrow m^{2} = 5 \Rightarrow m = \pm \sqrt{5}$
(iii) Given, $\frac{x^{2}}{3} - \frac{y^{2}}{2} = 1$ ………………(i)
$\frac{x^{2}}{a^{2}} + y^{2} = 1$ ………..(ii) & $x^{2} + y^{2} = 4$ ……….(iii)
from (i) & (iii)________ $y^{2} = \frac{2}{5} & x^{2} = \frac{18}{5}$
So, from (ii)______ $\frac{18}{5 a^{2}} + \frac{2}{5} = 1 \Rightarrow a^{2} = 6$
(iv) Line ____ $\frac{x}{e} + \frac{y}{e^{1}} = 1; \frac{1}{e^{2}} + \frac{1}{{(e^{1})}^{2}} = 1$
Since, it touches the ellipse;
So, $\frac{1}{\sqrt{\frac{1}{e^{2}} + \frac{1}{{(e^{1})}^{2}}}} = r \Rightarrow r = 1$

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