The abscissa of the point on the curve 3y = 6x β 5x ^{ 3 } , the normal at which passes through origin is:
(A)1Β
(B) 1/3
(C) 2Β
(D) 1/2
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Last updated at Dec. 4, 2021 by Teachoo
Question 1 The abscissa of the point on the curve 3y = 6x β 5x3, the normal at which passes through origin is: 1 (B) 1/3 (C) 2 (D) 1/2 Given curve 3π¦=6π₯β5π₯^3 Finding π π/π π π π/π π=πβππ^π Now, Slope of normal Γ Slope of tangent = β1 Slope of normal (m) = (βπ)/(πΊππππ ππ πππππππ) = (β1)/(ππ¦/ππ₯) = (βπ)/((π β π π^π)) Finding equation of Normal with Slope (βπ)/(π βππ^π ) & Passing through origin (0, 0) (π¦ β π¦_1)/(π₯ β π¦_1 ) = m (π¦ β π)/(π₯ β π) = (β1)/(2 β5π₯^2 ) (π )/(π ) = (βπ)/(π βππ^π ) Now, let normal and the curve passes through (h, k). Since (h, k) lies on normal So, (h, k) will satisfy the equation of normal π/β = (β1)/(2 β 5β^2 ) π/π = π/(ππ^π β π ) Since (h, k) lies on curve So, (h, k) will satisfy the equation of curve 3π=6ββ5β^3 3k = h (6 β 5β^2) π/β = (β (6 β5β^2))/(3β ) π/π = (π βππ^π)/(π ) On Solving (1) and (2) π/(ππ^π β π) = (π βππ^π)/π 3 = (6 β 5h2) (5h2 β 2) 3 = 6 (5h2 β 2) β 5h2 (5h2 β 2) 3 = 30h2 β 12 β 25h4 + 10h2 3 + 12 = β25h4 + 30h2 + 10h2 15 = β25h4 + 40h2 25h4 β 40h2 + 15 = 0 5h4 β 8h2 + 3 = 0 For simplicity, let h2 = z The equation becomes, 5z2 β 8z + 3 = 0 5z2 β 5z β 3z + 3 = 0 5z (z β 1) β 3 (z β 1) = 0 (5z β 3) (z β 1) = 0 Now, Either, (5z β 3) = 0 z = 3/5 Putting h2 = z h2 = 3/5 h = Β±β(π/π) which does not match with any option Or, (z β 1) = 0 z = 1 Putting h2 = z h2 = 1 h = Β± 1 But h = -1 is not in option β΄ h = 1 is correct. β΄ Abscissa = x β coordinate = h = 1 So, the correct answer is (A)