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)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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