SAT1000 - P896

This is also known as the "taxicab metric".

Say the foci are at $F_1(-a,0)$ and $F_2(a,0)$ , with $a>0$ , and the point on the locus is $P(x,y)$ . Then by definition, $|x-a|+|x+a|+2|y|=C$

Note that the locus will be symmetric about both the $x-$ and $y-$ axes. So to work out the shape, we can take both $x$ and $y$ positive.

When $0<x \le a$ , we have $2a+2y=C$ which is a straight line parallel to the $x-$ axis.

When $a<x$ , $2x+2y=C$ which is a straight line with negative gradient.

This is enough to see that the answer must be A .

This is an example of a topic called Taxicab Geometry , where you measure distance as if you were going there in a taxi in a city like NYC, where the blocks are all 90 degrees (you can't say that for any city). In this case, we would like to find an ellipse, which has foci $F_1$ and $F_2$ .

Ellipses can take a variety of shapes in Taxicab Geometry. They can be hexagons, octagons (I think), 45 degree turned squares, and, if the points are very close to each other, they can actually be a completely filled in square.

Using this, we can rule out the last two options. All that's left is just to check which one of the first two works.

In Option $B$ , the focal radii (is that what it's called?) is measured as $3$ when you consider the point $(1.5, 0)$ , but it is measured as $4$ when you consider the point $(1.5, 0.5)$ . Thus, it is not constant, so we can rule out Option $B$ .

We are left with our answer: $\boxed{\text{Option A}}$ .