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Q.

Let the foci of the hyperbola x2A2y2B2=1 be the vertices of the ellipse x2a2+y2b2=1 and the foci of the ellipse be the vertices of the hyperbola. Let the eccentricities of the ellipse and hyperbola be eE and eH, respectively. Then the match the following.

COLUMN - ICOLUMN - II
(A) bB is equal to(p) 1
(B) eH+eE is always greater than(q) 2
(C) If angle between the asymptotes of hyperbola is 2π3, then 4eE is equal to(r) 3
(D) If eE =12 and (x, y) is point of intersection of ellipse and the hyperbola then x2y2 is(s) 4

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a

A-p; B-q; C-q; D-s

b

A-p; B-q; C-s; D-q

c

A-p; B-q; C-s; D-s

d

A-s; B-q; C-q; D-s

answer is A.

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Detailed Solution

Equation of hyperbola x2A2y2B2=1

Vertices = (A, 0)

 Foci =AeH,0;

 Equation of ellipse x2a2+y2b2=1,

Vertex= (a, 0),Foci=(aeE,0)

Now From the given data AeH,0=(a,0) and aeE,0=(A,0)                                        

AeHaeE=AaeHeE=1

(A)bB=a21eE2A2eH21=a2A2a2A2=1

 (B) Now AMGMeH+eE>2

 (C) Angle between asymptotes 

2tan1BA=2π3 BA=3 eH=1+B2A2=2 eE=12 Now 4eE=2 

 D) Solve curves x2=2a2eE2b21eE2=92, x2y2=4

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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 of the hyperbola. Let the eccentricities of the ellipse and hyperbola be eE and eH, respectively. Then the match the following.COLUMN - ICOLUMN - II(A) bB is equal to(p) 1(B) eH+eE is always greater than(q) 2(C) If angle between the asymptotes of hyperbola is 2π3, then 4eE is equal to(r) 3(D) If eE =12 and (x, y) is point of intersection of ellipse and the hyperbola then x2y2 is(s) 4