$d = \sqrt{(x - 4a)^2 + (y - 0)^2}$

$d = \sqrt{(x - 4a)^2 + y^2}$

from

$a^2 y = x^3$

$y = \dfrac{x^3}{a^2}$

$y^2 = \dfrac{x^6}{a^4}$

$d = \sqrt{(x - 4a)^2 + \dfrac{x^6}{a^4}}$

$\dfrac{dd}{dx} = \dfrac{2(x - 4a) + \dfrac{6x^5}{a^4}}{2\sqrt{(x - 4a)^2 + \dfrac{x^6}{a^4}}} = 0$

$(x - 4a) + \dfrac{3x^5}{a^4} = 0$

$3x^5 + a^4 x - 4a^5 = 0$

by trial and error:

$x = a$

$y = \dfrac{a^3}{a^2} = a$

The nearest point is (a, a). *answer*