We draw two lines tangent to the ellipse parallel to the given line.
Their point of tangency will be the closest and farthest points
on the ellipse from the given line. 





Differentiate implicitly:


Solve for 



We find the slope of the line:





Find the points on the ellipse where the slope of the tangent lines
are parallel to the given line by setting the derivative of ellipse
equal to slope of line: 


solve for x
 

Substitute in ellipse's equation















Four points on the ellipse are (-4,2), (-4,-2), (4,2), (4,-2)

The only 2 points on the green tangent lines above are (-4,2) and (4,-2).

They are the required points.

Edwin

We can also do the problem without using calculus:



We draw two lines tangent to the ellipse parallel to the given line.
Their points of tangency will be the closest and farthest points
on the ellipse from the given line. 



We put the given line in slope-intercept form






The green tangent lines will have the same slope with a different
y-intercept b

So either green tangent line will have equation 

When we solve the system by substitution



There must be one and only one point of intersection.







For there to be only one solution for the point of tangency, the
discriminant must = 0









Substituting 






x=-4
Substituting in the equation of the line:




So one point is (-4,2), that's the closest point.

Substituting 






x=4
Substituting in the equation of the line:




So the other point is (4,-2), that's the farthest point.

Edwin