Trigonometric EquationHard
Question
A hyperbola, having the transverse axis of length 2 sinθ, is confocal with the ellipse 3x2 + 4y2 = 12. Then its equation is
Options
A.x2 cosec2 θ - y2 sec2 θ = 1
B.x2 sec2 θ - y2 cosec2 θ = 1
C.x2 sin2 θ - y2 cos2 θ = 1
D.x2 cos2 θ - y2 sin2 θ = 1
Solution
The given ellipse is 
= 1
⇒ a = 2, b = √3 ⇒ 3 = 4 (1 - ee) ⇒ e =
so that ae = 1
Hence the eccentricity e1, of the hyperbola is given by
1 = e1 sin θ ⇒ e1 = cosec θ
⇒ b2 = sin2θ(cosec2θ - 1) = cos2θ
Hence the hyperbola is
or x2cosec2θ - y2sec2θ = 1
= 1 ⇒ a = 2, b = √3 ⇒ 3 = 4 (1 - ee) ⇒ e =
so that ae = 1
Hence the eccentricity e1, of the hyperbola is given by
1 = e1 sin θ ⇒ e1 = cosec θ
⇒ b2 = sin2θ(cosec2θ - 1) = cos2θ
Hence the hyperbola is
or x2cosec2θ - y2sec2θ = 1Create a free account to view solution
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