Point on paraboloid at which the tangent plane is parallel to plane

Find pt on paraboloid $\displaystyle x = 5y^2 + 7z^2$, if it exists, at which the tangent plane is parallel to plane $\displaystyle -x + y + z = 3$.

Not completely sure how to approach this problem. I'm pretty sure it involves the gradient, so I set $\displaystyle f(x,y,z) = x - 5y^2 - 7z^2$ and found that gradient which was $\displaystyle \nabla f = i - 10yj - 14zk$.

Do I need the equation for the tangent plane of the paraboid? And are -1, 1, 1 the directional numbers for the plane $\displaystyle -x + y + z = 3$?

Re: Point on paraboloid at which the tangent plane is parallel to plane

Quote:

Originally Posted by

**deezy** Find pt on paraboloid $\displaystyle x = 5y^2 + 7z^2$, if it exists, at which the tangent plane is parallel to plane $\displaystyle -x + y + z = 3$.

Not completely sure how to approach this problem. I'm pretty sure it involves the gradient, so I set $\displaystyle f(x,y,z) = x - 5y^2 - 7z^2$ and found that gradient which was $\displaystyle \nabla f = i - 10yj - 14zk$.

Do I need the equation for the tangent plane of the paraboid? And are -1, 1, 1 the directional numbers for the plane $\displaystyle -x + y + z = 3$?

Two planes are parallel if their normal vectors are parallel.

The normal vector of the plane is $\displaystyle - \mathbf{i}+ \mathbf{j}+ \mathbf{k}$

or $\displaystyle \mathbf{i}- \mathbf{j}- \mathbf{k}$

Now just set this equal to the normal vector of the surface that you have found and find what the variables have to be.

$\displaystyle \nabla f = \mathbf{i} - 10y\mathbf{j} - 14z \mathbf{k} $

Re: Point on paraboloid at which the tangent plane is parallel to plane

Ok, here's what I've done:

$\displaystyle -10y = 1$

$\displaystyle -14z = 1$

$\displaystyle y = -1/10$

$\displaystyle z = -1/14$

I got confused at this point. I wasn't sure what I do with the x. $\displaystyle \nabla f = i -(10y)j-(14z)k$. There wasn't an x in the equation, and $\displaystyle 1 \neq -1$, so does the point exist?

And is there a way to check, if I found a point, that it was a correct point?

Re: Point on paraboloid at which the tangent plane is parallel to plane

Quote:

Originally Posted by

**deezy** Ok, here's what I've done:

$\displaystyle -10y = 1$

$\displaystyle -14z = 1$

$\displaystyle y = -1/10$

$\displaystyle z = -1/14$

I got confused at this point. I wasn't sure what I do with the x. $\displaystyle \nabla f = i -(10y)j-(14z)k$. There wasn't an x in the equation, and $\displaystyle 1 \neq -1$, so does the point exist?

And is there a way to check, if I found a point, that it was a correct point?

You have forgotten that every plane has two normal vectors, as I pointed out in my original post. If you muliply the normal by -1 you get get another normal vector but with opposite direction.