Find the point on the curve x^2/4 + y^2 = 1 which is as close as possible to the line y = 5 -x.

The hint on the problem says one way of minimize the distance between 2 curves is to pick generic points on both curves and the minimize the (square of ) the distant between 2 points.

I can't think of anything at all except the fact that the shortest distance between 2 curves are always along the line that is normal to both of them.

I have no idea where to start. Someone, Please, help me with this. Thank you!

2. Originally Posted by ka_rei
I can't think of anything at all accept the fact that the shortest distance between 2 curves are always along the line that is normal to both of them.
This is a good observation which allows an alternative solution (not the way where you pick two generic points, minimizing the distance).

The given line has slope -1, so the normal line has slope 1, making it of the form y = x + k. Now you want to know the point on the ellipse where this line intersects it perpendicularly.

Implicit differentiation yields that the tangent direct is -x/(4y). This is normal to the slope 1 if their product is -1, so: -x/(4y) = -1 <=> x = 4y. Substitution in the equation of the ellipse now gives a quadratic equation in x or y.