Let the foci of the hyperbola x2A2−y2B2=1 be the vertices of the ellipse x2a2+y2b2=1 and the foci of the ellipse be the vertices to the hyperbola. Let the eccentricities of the ellipse and hyperbola be eE and eH respectively, then match the followingColumn IColumn II(i)bB is equal to(A)1(ii)eH+eE is always greater then λ then least integral value of λ is(B)2(iii)If angle between the asymptotes of hyperbola is 2π3 then 4eE is equal to(C)3(iv)If eE2=12 and x,y is point of intersection of ellipse and the hyperbola, then x2y2 is(D)4 (E)5

We have A=aeE and a=AeH⇒eE.eH=1⇒eE+eH>2B2=A2eH2−1=a21−eE2=b2⇒bB=1Also, angle between the asymptotes is2tan−1BA=2π3BA=3⇒baeE=3⇒eE2=14Solving, x2a2+y2b2=1And x2a2eE2+y2b2=1We get x2y2=2a2eE2b21−eE2=4

Let the foci of the hyperbola x2A2−y2B2=1 be the vertices of the ellipse x2a2+y2b2=1 and the foci of the ellipse be the vertices to the hyperbola. Let the eccentricities of the ellipse and hyperbola be eE and eH respectively, then match the followingColumn IColumn II(i)bB is equal to(A)1(ii)eH+eE is always greater then λ then least integral value of λ is(B)2(iii)If angle between the asymptotes of hyperbola is 2π3 then 4eE is equal to(C)3(iv)If eE2=12 and x,y is point of intersection of ellipse and the hyperbola, then x2y2 is(D)4 (E)5

We have A=aeE and a=AeH⇒eE.eH=1⇒eE+eH>2B2=A2eH2−1=a21−eE2=b2⇒bB=1Also, angle between the asymptotes is2tan−1BA=2π3BA=3⇒baeE=3⇒eE2=14Solving, x2a2+y2b2=1And x2a2eE2+y2b2=1We get x2y2=2a2eE2b21−eE2=4

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Let the foci of the hyperbola $\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ be the vertices of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and the foci of the ellipse be the vertices to the hyperbola. Let the eccentricities of the ellipse and hyperbola be $e_{E}$ and $e_{H}$ respectively, then match the following
Column I
Column II
(i) $\frac{b}{B}$ is equal to (A)
1
(ii) $e_{H} + e_{E}$ is always greater then $λ$ then least integral value of $λ$ is (B)
2
(iii) If angle between the asymptotes of hyperbola is $\frac{2 π}{3}$ then $4 e_{E}$ is equal to (C)
3
(iv) If $e_{E}^{2} = \frac{1}{2}$ and $(x, y)$ is point of intersection of ellipse and the hyperbola, then $\frac{x^{2}}{y^{2}}$ is (D)
4
(E)
5

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$(i) - A, (i i) - B, (i i i) - B, (i v) - D$

$(i) - A, (i i) - B, (i i i) - B, (i v) - B$

$(i) - D, (i i) - C, (i i i) - A, (i v) - B$

$(i) - D, (i i) - A, (i i i) - B, (i v) - C$

answer is A.

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Detailed Solution

We have $A = a e_{E}$ and $a = A e_{H}$

$\Rightarrow e_{E} . e_{H} = 1$

$\Rightarrow e_{E} + e_{H} > 2$

$B^{2} = A^{2} (e_{H}^{2} - 1) = a^{2} (1 - e_{E}^{2}) = b^{2}$

$\Rightarrow \frac{b}{B} = 1$

Also, angle between the asymptotes is

$2 \tan^{- 1} \frac{B}{A} = \frac{2 π}{3}$

$\frac{B}{A} = \sqrt{3}$

$\Rightarrow \frac{b}{a e_{E}} = \sqrt{3}$

$\Rightarrow e_{E}^{2} = \frac{1}{4}$

Solving, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

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

We get $\frac{x^{2}}{y^{2}} = \frac{2 a^{2} e_{E}^{2}}{b^{2} (1 - e_{E}^{2})} = 4$

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