# Finding the parametric equations for tangent line to an ellipse at a given point

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.

• 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 $t=\pi/4$? Apr 18 at 2:09
• 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 $(z- 2) / - 2 = (x - 1) / 1 = (y-2) / 0 = t$ Apr 18 at 3:12

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 $$\boxed{x = t + 1, \ \ y = 2, \ \ z = - 2 t + 2}$$