(A) The locus of the point whose chord of contact with respect to the hyperbola x^2/16 - y^2/9 = 1

Question

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

Answer 1

(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 ⇒ 4√2

 Hence, the hyperbola is x² − y² = 32

 Answer: (D) → (r)