Here's a problem statement:
Find the minimum distance between a point $P(x,y)$ on the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ with $x,y\ge0$ and the line $x-y-5=0$.
It is clear that the line is a tangent to the ellipse because if we solve the line and ellipse we get equal roots. But the point of contact lies in the fourth quadrant. Therefore there is no common normal in the given arc of the ellipse and the line. So the method of finding the minimum distance between the ellipse and line with the help of common normal won't work here.
I took a point $(4|\cos\theta|, 3|\sin\theta|)$ on the ellipse. And I simply calculated the distance to the line, that is $$ d = \frac{|4|\cos\theta| - 3|\sin\theta| -5|}{\sqrt2}. $$ It is easy to show that $4|\cos\theta| - 3|\sin\theta|\in[-3, 4]$. So the minimum value of $d$ is $\boxed{\dfrac1{\sqrt2}}$.
My question: Is there any way to this geometrically without help of coordinates or trigonometry?
Thanks.
The ellipse is centered in origin $(0,0),$ its semin-major and semi-minor axis are $4$ and $3,$ respectively.
The line cuts the coordinate axes in points $(0,-5),(5,0).$
The ellipse is located above the line.
The point $(x,y)$ on the ellipse with $x,y\ge0$ and closest to the line is the vertex $A(4,0).$