Trigonometric EquationMedium
Question
Subnormal at any point ′P′ on the curve is same as subnormal at image of ′P′ in y = x on the same curve.
If curve lies in first quadrant and passes through (1,0), then -
If curve lies in first quadrant and passes through (1,0), then -
Options
A.Equation of curve can be x3/2 + y3/2 = 1
B.Tangent of slope 21/3 touches the curve at (41/3,1)
C.Tangent with equal intercepts on axes touches the curve at (41/3, 41/3)
D.If curve also passes through (0,1) then area bounded by the curve & axes is 
(sin 2θ)1/3dθ
(sin 2θ)1/3dθSolution
|yy′| = 
⇒
⇒
= c
∵ curve passes through (1,0)
(A) curve is x3/2 ± y3/2 = 1
(B) for x3/2 - y3/2 = 1 slope of tangent at point (41/3, 1) is 21/3
(C) (41/3,41/3) does not lie on the curve
(D) curve is x3/2 + y3/2 = 1
A =
(1 + x3/2)2/3 dx
put x3/2 = cos2θ
√x dx = - sin 2θ dθ
⇒ A =
⇒ A =
sin7/3 θ cos1/3 θdθ
A =
cos7/3 θ sin1/3 θdθ
⇒ 2A =sin1/3 θ cos1/3θ(sin2θ + cos2θ) dθ
A =
(sin 2θ)1/3dθ
A =
(sin 2θ)1/3 dθ
⇒
⇒
= c∵ curve passes through (1,0)
(A) curve is x3/2 ± y3/2 = 1
(B) for x3/2 - y3/2 = 1 slope of tangent at point (41/3, 1) is 21/3
(C) (41/3,41/3) does not lie on the curve
(D) curve is x3/2 + y3/2 = 1
A =
(1 + x3/2)2/3 dxput x3/2 = cos2θ
√x dx = - sin 2θ dθ⇒ A =
⇒ A =
sin7/3 θ cos1/3 θdθA =
cos7/3 θ sin1/3 θdθ⇒ 2A =
sin1/3 θ cos1/3θ(sin2θ + cos2θ) dθA =
(sin 2θ)1/3dθA =
(sin 2θ)1/3 dθCreate a free account to view solution
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