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

Answer

25

Explanation

The centre and the radius of the given circle are P(6,0) and `sqrt(5)`, respectively. Now, the minimum distance between the two curves always occurs along a line which is normal to both the curves.
The equation of the normal for `y^(2)=4x` at (`t^(2),2t`) is
`y=-tx+2t+t^(3)`.
If it is normal to the circle also, then it must pass though (6,0). So
`0=t^(3)-4trArr t = 0 ` or ` t = +-2`
`rArr A (4,4)` and `C(4,4)`
`rArr PA=PC=sqrt(20)=2sqrt(5)`
The required minimum distance is `2sqrt(5)-sqrt(5)=sqrt(5)=d`
Hence `d^(4)=(sqrt(5))^(4)=25`