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Answer 1 · 2018-02-12T15:03:58

#d= 1#

The generic point lying on the parabola of equation:

#y^2 = 4x#

has coordinates: #P(y^2/4, y)#.

The distance between such point and the point #P(1,0)# is:

#d(y) = sqrt((y^2/4-1)^2+y^2) = sqrt( (y^2-4)^2/16 +y^2) = sqrt(y^4+8y^2+16 )/4#

To find the shorter distance we can then minimize:

#f(y) = y^4+8y^2+16#

#(df)/dy = 4y^3 +16y = 0#

#y(y^2+4) = 0#

and the only solution is #y=0#.

#(d^2f)/dy^2 = 12y^2+16#

and as for #y= 0# the second derivative is positive the point is a relative minimum.

Then the shorter distance is:

#d(0) = 1#

graph{y^2-4x=0 [-10, 10, -5, 5]}