Find the shortest distance of the point `(0, c)` from the parabola `y=x^2,w h e r e0lt=clt=5.`

Answer

`sqrt(c - (1)/(4)), (1)/(2) le c le 5`

Explanation

Let the point Q (`x, x^(2))` on `x^(2) = y` whose distance from (0,c) is minimum.
Now `PQ^(2) = x^(2) + (x^(2) - c)^(2)`
Let `f(x) = x^(2) + (x^(2) - c)^(2)`
`f'(x) = 2x + 2 (x^(2) - c) 2x`
`= 2x (1 + 2x^(2) - 2c) = 4x (x^(2) - c + (1)/(2))`

`4x (x - sqrt(c - (1)/(2))) (x + sqrt(c - (1)/(2)))`, when `c gt (1)/(2)`
For maxima put `f' (x) = 0`
Now, `f''(x) = 4 [x^(2) - c + (1)/(2)] + 4x [2x]`
At `x = +- sqrt(c - (1)/(2))`
`f'' (x) ge 0`
`:. f(x)` is minimum
Hence, minimum value of `f(x) = |PQ|`
`sqrt((sqrt(c - (1)/(2)))^(2) + ((srt(c - (1)/(2)))^(2) - c)^(2))`
`= sqrt(c - (1)/(2) + (c + (1)/(2) - c)^(2)) = sqrt(c - (1)/(4)), (1)/(2) le c le 5`