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in Mathematics by (53.1k points)

Match the items of Column I with those of Column II.

Column I Column II
(A) The locus of the point whose chord of contact with respect to the hyperbola x2/16 - y2/9 = 1 touches the circle described on the line joining the foci is (p) (x2 + y2) = 4x2 - 3y2
(B) The chords of the circle x2 + y2 = 4 touch the hyperbola x2/4 - y2/3 = 1. Then the locus of the midpoints of these chords is (q) x2 + y2 = 9
(C) The director circle of the hyperbola x2/25 - y2/16 = 1 is (r)  x2 - y2 = 32
(D) The distance between the foci of a hyperbola is 16 and its eccentricity is √2. Then the equations of the hyperbola is  (s) x2/256 + y2/81 = 1/25
(t) x2 - y2 = 64

1 Answer

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(A) P(x1, y1) be a point on the locus. That is xx1/16 - yy1/9 = 1 touches the circle described on the line joining the foci S(5, 0) and S'(−5, 0) whose equation is

Therefore, the locus is

x2/162 + y2/92 = 1/25

Answer: (A)  (s)

(B) (x1, y1) is the midpoint of a chord x2 + y2 = 4 touching the hyperbola x2/4 - y2/3 = 1

That is, the line xx1 + yy1 = x2 1 + y21 touches the hyperbola. This means

Therefore, the locus is

(x2 + y2)2 = 4x - 3y2

Answer: (B)  (p)

(C)  Director circle of

x2/a2 - y2/b2 = 1

is x2 + y2 = a2 − b2. Hence, a2 = 25 and b2 = 16. Hence, the director circle is x2 = y2 = 9

 Answer: (C)  (q)

(D)  Since 2 is the eccentricity of the hyperbola, it must be a rectangular hyperbola. Hence, it is of the form x2 − y2 = a2. By hypothesis,

 2ae = 16 ⇒2a(2)16 ⇒ 42

 Hence, the hyperbola is x2 − y2 = 32

 Answer: (D)  (r)

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