Locus of boundary when shadow is taken

Question

For the first part (i), I could solve by taking images and I got the answer as ellipse, but for the second part I don't know how to take shadow.

Can I get exact equation of locus or do I get just the name of locus? The answer is parabola.

Answer 1

Let $A(5,0,3)$. The equation of the sphere is $$(x-2)^2+(y-0)^2+(z-2)^2=1\tag1$$

Let $P(X,Y,0)$ be a point on the $xy$ plane. Now, let us define $Q$ as a point where the line $AP$ is tangent to the sphere.

Then, since $Q$ is on the line $AP$, we can write, using a parameter $t$,
$$\vec{OQ}=\vec{OA}+t\vec{AP}=(tX-5t+5,tY,-3t+3)$$

So, from $(1)$, we get $$(tX-5t+5-2)^2+(tY-0)^2+(-3t+3-2)^2=1,$$ i.e. $$(X^2-10X+Y^2+34)t^2+(6X-36)t+9=0\tag2$$ which is a quadratic equation on $t$.

Since $Q$ is a point where the line $AP$ is tangent to the sphere, the discriminant of $(2)$ has to be $0$, so we get $$(6X-36)^2-4\times(X^2-10X+Y^2+34)\times 9=0,$$ i.e. $$Y^2=-2X+2$$ which is the equation of a parabola.