Q.228. Points of tangency and radius of curvature.

Question 228.
Find the co-ordinates of real points on the curve y^2 = 2x (3 - x^2), the tangents at which are parallel to    x-axis. Show that the radius of curvature at each of these point is 1/3.

Answer 228.
y^2 = 2x (3 - x^2)
=> y^2 = 6x - 2x^3
=> 2y dy/dx = 6 - 6x^2

For tangents parallel to the x-axis, dy/dx = 0
=> 6 - 6x^2 = 0
=> x = -1 or 1
For x = - 1, y is not in R
For x = 1, y = -2 or 2
=> the points are (1, -2) and (1, 2).

dy/dx = (3 - 3x^2)/y
=> d^2y/dx^2 = [-6xy - (3 - 3x^2) dy/dx] / y^2
=> d^2y/dx^2 at (1, -2) and (1, 2) = ±3

Radius of curvature to the curve y = f(x) at any point is given by
[1 + (dy/dx)^2]^(3/2) / l d^2y/dx^2 l
= [1 + 0]^(3/2) / l ±3 l
= 1/3.

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