# Thread: Do I need to change this into parametric form?

1. ## Do I need to change this into parametric form?

I need to find the points on:

$x^2+\frac{y^2}{4}+\frac{z^2}{9}=1$

where the tangent line is parallel to:

$x+2y+z=0$

This is what I've done:

$\langle 2x, \frac{y}{4}, \frac{2z}{9} \rangle \cdot \langle 1,2,1 \rangle =0$

$=2x+y+\frac{2z}{9}=0$

Not sure what to do now... do I have to change this to parametric form? Should I have changed it first?

2. Originally Posted by MathSucker
I need to find the points on:

$x^2+\frac{y^2}{4}+\frac{z^2}{9}=1$

where the tangent line is parallel to:

$x+2y+z=0$

This is what I've done:

$\langle 2x, \frac{{\color{red}2}y}{4}, \frac{2z}{9} \rangle \cdot \langle 1,2,1 \rangle =0$
If the planes are parallel then their normal vectors should also be parallel, not perpendicular. So your equation should be $\langle 2x, \tfrac{y}{2}, \tfrac{2z}{9} \rangle = \lambda \langle 1,2,1 \rangle$. Then $\langle x,y,z\rangle = \langle\tfrac\lambda2,4\lambda,\tfrac{9\lambda}2\r angle$. Substitute those values for x, y, z into the equation $x^2+\frac{y^2}{4}+\frac{z^2}{9}=1$ to find the two values of $\lambda$ for which that point lies on the ellipsoid.

3. So if I get $\lambda= \pm \frac {2}{\sqrt{13}}$,

are the two points:

$\pm (\frac{2}{\sqrt{13}}, \frac{4}{\sqrt{13}}, \frac{2}{\sqrt{13}})$