A light source is at O . Find the contour of the shadow of…

A light source is at $O$. Find the contour of the shadow of the sphere $x^2+y^2+z^2-10x+2y+10=0$ on the plane $3x-2y=30$.

Explanation of Method

The shadow contour of a sphere lit from the origin on a vertical plane.

Step 1: Tangent cone $\cap$ plane

The contour is the intersection of the cone $15x^2-10xy-9y^2-10z^2=0$ (Problem 317) with the plane. Substituting $y=\tfrac{3x-30}2$: $15x^2-5x(3x-30)-\tfrac94(3x-30)^2-10z^2=0\iff 81x^2-2220x+40z^2+8100=0$.

\[81x^2-2220x+40z^2+8100=0,\qquad 3x-2y=30\]

Step 2: Shape

\[ 81\left(x-\tfrac{370}{27}\right)^2+40z^2=\tfrac{576000}{81}\iff \frac{\left(x-\frac{370}{27}\right)^2}{\frac{64000}{729}}+\frac{z^2}{\frac{14400}{81}}=1 . \]

This is an ellipse with center $\left(\tfrac{370}{27},\ \tfrac{50}9,\ 0\right)$. The vertical semi-axis is $\tfrac{40}3$. In the $x$-direction the half-extent is $\tfrac{80\sqrt{10}}{27}\approx9.37$.

The plane is tilted: along it, $y$ changes by $\tfrac32$ per unit of $x$. So the true horizontal semi-axis, measured in the plane along $\{2,3,0\}$, is

\[ a=\frac{80\sqrt{10}}{27}\cdot\frac{\sqrt{13}}{2}=\frac{40\sqrt{130}}{27}\approx16.89,\qquad b=\frac{40}3\approx13.33 . \]

$O$ $\left(\tfrac{370}{27},\tfrac{50}9,0\right)$ $3x-2y=30$

Light from $O$: the tangent cone (orange rays) cuts the plane $3x-2y=30$ in the shadow ellipse (dark), with semi-axes $\tfrac{40\sqrt{130}}{27}\approx16.9$ and $\tfrac{40}3\approx13.3$.

Result

\[\boxed{\,81x^2-2220x+40z^2+8100=0,\qquad 3x-2y=30\,}\]

Remark

Correction: the source calls $\tfrac{80\sqrt{10}}{27}$ the semi-axis. That value is only its projection onto the $x$-axis. The real semi-axis of the shadow ellipse is $\tfrac{40\sqrt{130}}{27}$.

Also note: the distance from $C(5,-1,0)$ to the plane is $\tfrac{|15+2-30|}{\sqrt{13}}=\sqrt{13}\approx3.61\lt 4$. So the sphere actually cuts through the "screen". The contour above is still the boundary of the cone of rays that graze the sphere.