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प्रश्न
If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that `[1+ (dy/dx)^2]^(3/2)/((d^2y)/dx^2)` is a constant independent of a and b.
उत्तर
Here, (x – a)2 + (y – b)2 = (Given) …(1)
On differentiating with respect to x,
`=> 2 (x - a) + 2(y - b)^2 dy/dx = 0`
`=> (x - a) + (y - b) dy/dx = 0` ...(2)
Differentiating again with respect to x,
`1 + dy/dx * dy/dx + (y - b) (d^2 y)/dx^2` = 0
`1 + (dy/dx)^2 + (y - b) (d^2y)/dx^2` = 0
`=> (y - b) = - {(1 + (dy/dx)^2)/((d^2y)/dx^2)}` ...(3)
Putting the value of (y – b) in (2),
`(x - a) = (1 + (dy/dx)^2)/((d^2 y)/dx^2) * dy/dx`
या `(x - a) = {(1 + (dy/dx)^2)/((d^2y)/dx^2)}(dy/dx)` ...(4)
Putting the values of (x - a) and (y - b) from (3) and (4) in (1),
`{1 + (dy/dx)^2}^2/((d^2y)/dx^2)^2 * (dy/dx)^2 + {(1 + (dy/dx)^2)/((d^2y)/dx^2)} = c^2`
On multiplying by `((d^2y)/dx^2)^2,`
`[1 + (dy/dx)^2]^2 (dy/dx)^2 + [1 + (dy/dx)^2]^2`
`= c^2 ((d^2y)/dx)^2`
`=> [1 + (dy/dx)^2]^2 [(dy/dx)^2 + 1] = c^2 ((d^2y)/dx^2)^2`
`=> {1 + (dy/dx)^2}^3 = c^2 ((d^2y)/dx^2)^2`
On taking the square root,
`therefore {1 + (dy/dx)^2}^(3//2)/((d^2y)/dx^2)` = c ...(a constant independent of a and b.)
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