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The original question is:

The ellipsoid $4x^2+2y^2+z^2=16$ intersects the plane $y=2$ in an ellipse. Find parametric equations for the tangent line to this ellipse at the point $(1,2,2)$.

I know the slope is $-2$ and the equation of the tangent line at point $(1,2,2)$ is $z-2=-2(x-1), y=2$. The given solution jumped directly from here to the conclusion.

The problem is that I have no idea how to get the directional vector of this tangent line.

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    $\begingroup$ I've always found it helpful to express these intersections in set$-$builder notation: $$\begin{eqnarray*}\{(x,y,z):4x^2+2y^2+z^2=16,y=2\} &=& \{(x,2,z):4x^2+z^2=8\} \\ &=& \Big\{\left(\sqrt{2}\cos(t),2,\sqrt{8}\sin(t)\right):t\in [0,2\pi)\Big\}\end{eqnarray*}$$ Evidently $\vec{r}(t)=\left(\sqrt{2}\cos(t),2,\sqrt{8}\sin(t)\right):t\in[0,2\pi)$ is a parametric representation of your intersection. Can you find equation of tangent line to $\vec{r}(t)$ at $t=\pi/4$? $\endgroup$
    – Matthew H.
    Apr 18 at 2:09
  • $\begingroup$ Given the ellipse is in the plane $y = 2$, the direction vector will also be in the plane $y = 2$ and so the y-component of the direction vector is simply zero. The equation of the plane that you have written can be simply rewritten as $(z- 2) / - 2 = (x - 1) / 1 = (y-2) / 0 = t$ $\endgroup$
    – Math Lover
    Apr 18 at 3:12

1 Answer 1

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We write down the equations of the ellipsoid and the plane: $$ 4 x^2 + 2 y^2 + z^2 = 16 \tag{1} $$ $$ y = 2 \tag{2} $$

When (1) and (2) intersect, we find that $$ 4 x^2 + 2 (4) + z^2 = 16 \ \ \mbox{or} \ \ 4 x^2 + 8 + z^2 = 16 $$

Thus, the intersection of (1) and (2) is an ellipse: $$ 4 x^2 + z^2 = 8 \tag{3} $$

We know that the tangent line on the ellipsoid at the point $(1, 2, 2)$ must be an element of the plane $y = 2$.

From the equation of the ellipse, we find the slope at the point $(1, 2)$ using the equation (3): $$ 8 x + 2 z z_x = 0 $$ or $$ z_x = - {8 x \over 2 z} = - {4 x \over z} $$

At the point $(1, 2)$, $$ z_{x} = - {4 \over 2} = -2 $$

Now, the equation of the tangent line to the ellipse (3) at the point (1, 2) is $$ z - 2 = m (x - 1) = -2 (x - 1) $$ since $m = -2$ is the slope.

Simplifying, we get the tangent line as $$ z = - 2 x + 4 \tag{4} $$

This shows that the tangent line for the ellipsoid (1) at the point (1, 2, 2) is: $$ z = - 2 x + 4, \ \ y = 2 $$

We recall that the parametric equations for a line passing through a point $(x_0, y_0, z_0)$ in the direction of the vector $(l, m, n)$ in the 3-D space is: $$ {x - x_0 \over l} = {y - y_0 \over m} = {z - z_0 \over n} $$

It is given that $(x_0, y_0, z_0) = (1, 2, 2)$ and $m = 0$ (Note that the line lies in the plane $y = 2$).

From the equation of the tangent line $z = - 2 x + 4$ obtained in Eq. (4), we get the equation of the tangent line as $$ {x - 1 \over l} = {y - 2 \over 2} = {z - 2 \over 2} = t $$

Thus, $$ x = 1 + l t $$ $$ y = 2 $$ $$ z = 2 + n t $$ $$ 2 + n t = - 2 ( 1 + l t) + 4 $$ Simplifying, we get $$ 2 + n t = -2 - 2 l t + 4 $$ or $$ n t + 2 l t = (n + 2 l) t = 4 - 4 = 0 $$

This gives $$ n = - 2 l $$

If we take $l = 1$, then we get $n = -2$.

Thus, we write the parametric equations of the tangent line as $$ {x - 1 \over 1} = {y - 2 \over 0} = {z - 2 \over -2} = t $$ or equivalently $$ \boxed{x = t + 1, \ \ y = 2, \ \ z = - 2 t + 2} $$

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