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

Explanation

Let `A(sqrt(alpha),alpha)` be any point on the parabola ` y=x^(2)`
Then distance between A and (0,C) will be given by
`D(alpha)=sqrt(alpha+(alpha-C)^(2))`
Squaring both side
`D^(2)alpha=alpha+(alpha-C)^(2)`
Differentiating with respect to `alpha `
`2D(alpha). D'(alpha)=1+2(alpha-C)`
For maxima and minima
`D'(alpha)=0`
`2 alpha -2C+1=0`
`2alpha=2C-1 `
`alpha=(2C-1)/(2)`
Using first derivative test
`0 le C le 5 , " " alpha lt (2C-1)/(2) " "D'(alpha) lt 0`
` 0 le C le 5 " " alpha gt (2C-1)/(2) " "D'(alpha) gt 0`
So, ` alpha=(2C-1)/(2)` is point of minima
Hence, the required shortest distance is given by
`D(alpha)=(sqrt(4C-1))/(2)`