Finding point on paraboloid surface given normal line point

[Emethyst]

Homework Statement

[/B]
Find the coordinates of the point P on the surface of the paraboloid z=6x²+6y²-(35/6) where the normal line to the surface passes through the point (25/6, (25√22)/6, -4). Note that a graphing calculator may be used to solve the resulting cubic equation.

Homework Equations

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Gradient of the paraboloid ∇ƒ

The Attempt at a Solution

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This question has been stumping me all day as I don't know how to go about finding the point, even though it seems like it should be a piece of cake. I already understand from class that the gradient gets me the normal line slope which is ∇ƒ = <12x, 12y, -1> (where ƒ is the paraboloid function) and from this and the given point can form the parametric version of the normal line like so:

n = (25/6) + 50t, (25/6)√22 + (50√22)t, -4 - t

It's after this part where I'm lost, it seems from here I need to find t to finally solve for the 3 coordinates which from what I remember of linear algebra can be done by inputting the normal vector into the paraboloid function, yet the note about having to solve a cubic equation leaves me confused as doing it this way only ends up with an ugly quadratic.

If anyone can point me in the right direction it would be greatly appreciated as I'm pretty sure I'm missing something important here.

 

[Ray Vickson]

Homework Statement

[/B]
Find the coordinates of the point P on the surface of the paraboloid z=6x²+6y²-(35/6) where the normal line to the surface passes through the point (25/6, (25√22)/6, -4). Note that a graphing calculator may be used to solve the resulting cubic equation.

Homework Equations

[/B]
Gradient of the paraboloid ∇ƒ

The Attempt at a Solution

[/B]
This question has been stumping me all day as I don't know how to go about finding the point, even though it seems like it should be a piece of cake. I already understand from class that the gradient gets me the normal line slope which is ∇ƒ = <12x, 12y, -1> (where ƒ is the paraboloid function) and from this and the given point can form the parametric version of the normal line like so:

n = (25/6) + 50t, (25/6)√22 + (50√22)t, -4 - t

It's after this part where I'm lost, it seems from here I need to find t to finally solve for the 3 coordinates which from what I remember of linear algebra can be done by inputting the normal vector into the paraboloid function, yet the note about having to solve a cubic equation leaves me confused as doing it this way only ends up with an ugly quadratic.

If anyone can point me in the right direction it would be greatly appreciated as I'm pretty sure I'm missing something important here.

If the point P has coordinates $(x_0,y_0,z_0)$, the normal line through that point has equation
$$(x,y,z) = (x_0,y_0,z_0) + t (12 x_0,12 y_0,-1 ),$$
where $t$ is a scalar. That line must contain the point $(25/6, 25 \sqrt{22}/6, -4)$. In addition, $(x_0,y_0,z_0)$ must lie on the paraboloid.

 

[Dick]

The point you might be missing is that the given point does not lie on the paraboloid.

 

[Emethyst]

Many thanks for the help you guys, I know where I went wrong now, I messed up inputting the given point into the normal line equation it seems. Finally got the cubic function and an answer that makes sense.

Really appreciated it

 

[Sam3313]

Hey. I'm stuck on the same question. Can you please explain to me how to go from there? Thanks a ton

 

[Dick]

Hey. I'm stuck on the same question. Can you please explain to me how to go from there? Thanks a ton

You have to show us your own attempt to solve it first.

 

[Sam3313]

ok so here is my attempt.(sorry for the late reply) :)
The question: find the coordinates of point p on the surface of the paraboloid z=5(x^2+y^2)-(109/5) where the normal line to the surface goes through the point (41/5,(41/5)√ 38,-18)
my attempt:
so first i got the normal by using the gradient of the given equation of the paraboloid and got (10x,10y,-1) which with the point plugged in becomes (82,82√38,-1)
Then i thought that the normal line equation would be helpful for the next step which essentially is finding a point of intersection between the paraboloid and the normal but i didn't know what to do from there :/
equation of normal line
(x,y,z).(82,82√38,-1)=(82,82√38,-1).(41/5,(41/5)√ 38,-18)
and also tried the parametric form but still was confused as to what to do next:
parametric form:
n = (41/5,(41/5)√ 38,-18)+t(82,82√38,-1)
Thank you again for the support.