Let $x^2+y^2+(z-1)^2=1$ be a sphere in $\Bbb R^3$. $z=0$ be as a ground. Suppose a light source is located at $(d,0,d)$ ($d>0$). As $d\to\infty$, it is assumed that the light can be view as "parallel" light. Find the equation of the shade on the ground.
I have totally no idea about this. (I'd tried use the slanted cylinder equation, or tangent plane to the spheres, but still stuck on it. )Need help!
Consider the intersection of your sphere with the $XZ$-plane. This is a circle of radius $1$ tangent to the $X$ axis. The two rays of light coming from infinity under a $45$ degrees angle are tangent to that circle at the points $(\sqrt{2}/2,0,1-\sqrt{2}/2)$ and $(-\sqrt{2}/2,0,1+\sqrt{2}/2)$ (just draw a picture to convince yourself). Those two lines have direction vector $(1,0,1)$ so their equations are $$ (x,y,z)=(\sqrt{2}/2,0,1-\sqrt{2}/2)+t(1,0,1),\quad (x,y,z)=(-\sqrt{2}/2,0,1+\sqrt{2}/2)+t(1,0,1) $$ They intersect the $XY$ plane at $(\sqrt{2}-1,0,0)$ and $(-\sqrt{2}-1,0,0)$. Those are the endpoints of the major axis, so the center is at $(-1,0,0)$ and $a=\sqrt{2}$. Clearly, $b=1$ (no deformation in that direction). The equation of the shadow on the $XY$-plane is $$ \frac{(x+1)^2}{2}+y^2=1 $$