Let the point be ( x , x^2 -3)
Using the square of the the distance formula, we have
D^2 = (x -0)^2 + ( x^2 - 3 - 0)^2 simplify
D'2 = x^2 + x^4 - 6x^2 + 9
D^2 = x^4 - 5x^2 + 9
Take the derivative of D^2 and set to 0
4x^3 - 10x = 0 factor
x ( 4x^2 - 10) = 0
The second factor set to 0 gives us what we need
4x^2 - 10 = 0
2x^2 - 5 = 0
2x^2 = 5
x^2 = 5/2 due to symmetry.....there will be two points of equal distance...so....we can take the + root
x = sqrt (5/2)
And y = X^2 - 3 = 5/2 -3 = -1/2
So the point is (sqrt (5/2) , -1/ 2 )
And the disance is sqrt ( 5/2 + (-1/2)^2 ) = sqrt ( 11/4)
See the graph here : https://www.desmos.com/calculator/47lpdp6t98