Find the minimum distance between the curves `y^2=4xa n dx^2+y^2-12 x+31=0`

Explanation

We have parabola `y^(2)=4x` and `x^(2)+y^(2)-1x+31=0`
Shortest distance between two curves occur along common normal line.
Normal to circle always passes through its centre.
So, we have to find the normal to parabola which gose through centre of the circle.
Given circle is `(x-6)^(2)+y^(2)=5`.
So, centre is T(6,0) and radius is `sqrt(5)`.
Equation of normal to parabola having slope m is
`y=mx-2m-m^(3)`
If it passes through (6,0), then `0=6m-2m-m^(3)`
`:." "m^(3)-4m=0`
`:." "m=0,pm2`.
So, equation of normal and corresponding point on the parabola are tabulated below.

`:." "TQ=TR=sqrt(20)`
TP=6
So, required shortest distance `=sqrt(20)-sqrt(5)=sqrt(5)`