Find the minimum distance from the point (4,2) to the parabola y-8x.
Ans. 2V2 units
y-8x is not a parabola.
As oobleck said, you don't have the equation of a parabola, just a
mangled expression.
However, this would be a nice Calculus question for a real parabola.
I will assume your parabola is y = 8 - x^2, and your point outside is (4,2)
Using the property, that at the point of shortest distance to (4,2) the tangent would be perpendicular to the normal at that point, ....
let the point yielding the shortest distance be (a,b) or (a, 8 - a^2)
then slope of normal = (b-2)/(a-4)
dy/dx = -2x, and at (a,b) the slope of the tangent is -2a
and then (b-2)/(a-4) = 1/(2a), but b = 8-a^2, then ...
(8-a^2 - 2)/(a-4) = 1/(2a)
(6-a^2)/(a-4) = 1/(2a)
12a - 2a^3 = a-4
2a^3 - 11a - 4 = 0
(not a nice answer, but hopefully once you have fixed your parabola
it will solve nicely)
solve for a, then in b = 8y-12 you can find b and
you have your point.
(Once you fix your equation for the parabola, follow this method)