Question 1200122
<pre>
{{{drawing(400,300,-10,10,-5,10,
graph(400,300,-10,10,-5,10),

arc(0,0,12,12/sqrt(5)), line(-15,0,15,12) )}}}

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. 

{{{drawing(400,300,-10,10,-5,10,
graph(400,300,-10,10,-5,10),green(line(-14,-2,16,10),line(-16,-10,14,2)),

arc(0,0,12,12/sqrt(5)), line(-15,0,15,12) )}}}

{{{x^2 + 5y^2 = 36}}}

Differentiate implicitly:

{{{2x + 10y*expr(dy/dx)=0}}}
Solve for {{{expr(dy/dx)}}}

{{{dy/dx=-x/(5y)}}}

We find the slope of the line:

{{{2x − 5y + 30 = 0}}}
{{{y=expr(2/5)x+6}}}
{{{slope=2/5}}}

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: 

{{{x/(5y)=2/5}}}
solve for x
{{{x=-2y}}} 

Substitute in ellipse's equation

{{{x^2 + 5y^2 = 36}}}
{{{(-2y)^2+5y^2=36}}}
{{{4y^2+5y^2=36}}}
{{{9y^2=36}}}
{{{y^2=4}}}
{{{y="" +- 2}}}

{{{x^2 + 5y^2 = 36}}}
{{{x^2 + 5("" +- 2)^2=36}}}
{{{x^2 + 5(4)=36}}}
{{{x^2 +20=36}}}
{{{x^2=16}}}
{{{x="" +- 4}}}

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</pre>