1.  Our curve is specific !  If (x,y) is the point on the curve then

    2y^2 - x^2 = 8,  which implies  y^2 = .   (1)



2.  The distance from ANY point (x,y) on the coordinate plane to the point (3,0) is

     = .    (2)

    Now, if the point lies on the hyperbola, you have (1), and you can substitute this expression for  into the formula (2).

    You will get

     =  +  =  +  = .   (3)

    So, you need to minimize (3). In other words, you need to find the value of "x" which minimizes this quadratic function.



4.  The quadratic function of the general form q(x) =  achieves the maximum at x = .

    In your case this "x" is x = -  =  = 2.


    So, we found the value of "x". It is x= 2.

    Then the corresponding value of "y" on your curve is 

          = 8  ---->   = 8 + 4 = 12  ---->   =  = 6.

     Thus your "closest" point on the curve is  (x,y) = (,).



5.  Now you can find that minimal distance:

     =  =  = 1 + 6 = 7.


    Hence, the minimal distance itself is   = 2.646 (approximately).