Length of shortest path

Question

What is the length of shortest path that begins at the point (-1,1) , touches the x axis and then ends at point on the parabola $(x-y)^2 =2(x+y-4)$ .

I put $x=x-y$

$y=x+y-4$

then the starting point would be $(-4,-2)$ .

can I proceed by this methos of changing the coordinates

After that I got stuck. Can anybody help me?

Answer 1

Note that the parabola lies entirely in the first quadrant.

Let $P_1$ be the point $(-1,1)$ and $P_2$ be its reflectional image in the $x$-axis, i.e. $P_2=(-1,-1)$. Suppose that the path meets the $x$-axis at $Q$. The shortest path should consist of a line segment joining $P_1$, $Q$ and a line segment joining $Q$ and some point on the parabola. The two line segments are not in the same direction. But the line segment $P_1Q$ has the same length as $P_2Q$. So we can replace the starting point by $P_2$ and the path can be a line segment. So the shortest length is the shortest distance between $P_2$ and the parabola, which is equal to the distance between $P_2$ and the point $(2,2)$. So the length of the shortest path is $3\sqrt{2}$.

Answer 2

Your transformation maps $(0,0)$ to $(0,-4)$ and $(-1,1)$ to $(-2,-4)$. It does not preserve distances.