Find the points where the line tfrac x-2=tfrac y0=(z-3)/(-3)…

Find the points where the line $\tfrac x{-2}=\tfrac y0=\tfrac{z-3}{-3}$ meets the ellipsoid $\tfrac{x^2}4+y^2+\tfrac{z^2}9=1$.

Explanation of Method

Piercing points of a line and an ellipsoid.

Step 1: Substitute

$x=-2t,\ y=0,\ z=3-3t$: \[t^2+(1-t)^2=1\iff 2t(t-1)=0\iff t=0\ \text{ or }\ t=1 .\]

\[T_1(0,0,3),\qquad T_2(-2,0,0)\]

Checks: $\tfrac99=1$ and $\tfrac44=1$. $\checkmark$ Both points are vertices of the ellipsoid, and the line is the chord joining them in the plane $y=0$.

$T_1(0,0,3)$ $T_2(-2,0,0)$

The line lies in the plane $y=0$, which cuts the ellipsoid in $\tfrac{x^2}4+\tfrac{z^2}9=1$. The line meets this ellipse at the ends of two semi-axes.

Result

\[\boxed{\,T_1(0,0,3),\qquad T_2(-2,0,0)\,}\]