access 2013 Set Size and location of a pop-up form

I have an access 2013 application where I have created a pop-up reference form that is opened when I get focus on a field and then closed when I lose focus - The form is a datasheet view and I would like to size the form and set its location - is there a way to do this?
rogerdjrAsked:
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Dale FyeCommented:
In the forms Load event, use the move method to reposition and size the form, something like:

Private Sub Form_Load()

    Me.Move 3000, 3000, 7500, 5000
   
End Sub
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rogerdjrAuthor Commented:
I created this code to run on the got focus event of a control in the working form, opening a "reference form"

    DoCmd.OpenForm "CASp_zLkupImplementationPlanPriorityListFrm", acFormDS
    DoCmd.MoveSize -10080, 6480, 4320, 2160


which partially addresses the issue - two things I want to correct but I'm not sure how to do it

1) Set position relative to the control that currently has focus rather than to the available screen area (that is set the top of the reference form 1" right and at the same vertical position as the control on the working form)
2) I don't want the cursor to jump to the pop-up its just ref info.

thanks
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Dale FyeCommented:
You can use the form WindowTop and WindowLeft properties to get the relative position of the top left corner of a form.  Then, within the form, you have determine the relative position of the particular control you are working with, so to get the bottom of your current control, you might use something like:

me.windowTop + me.Detail.Height + me.Controls("yourControl").top + me.controls("yourControl").height

And the right side of that control might be:

me.WindowLeft + me.Controls("yourControl").Left + me.Controls("yourControl").width
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