Tangent Plane Problem: Where am I wrong?

[gikiian]

1. Homework Statement :

Find the points on the hyperboloid of two sheets with equation x²-2y²-4z²=16 at which, the tangent plane is parallel to the plane 4x-2y+4z=5.

Homework Equations

The hyperboloid with two sheets: x²-2y²-4z²=16

The given plane: 4x-2y+4z=5

Equation of tangent plane: Fx(x-x₀) + Fy(y-y₀) + Fz(z-z₀) = 0 ;
Fx, Fy and Fz are partial derivatives
3. The Attempt at a Solution :

For the hyperboloid, we can say that

F(x,y,z)=> x² - 2y² - 4z² - 16= 0​

Equation of the tangent plane is given by:

Fx(x-x₀) + Fy(y-y₀) + Fz(z-z₀) = 0​

so we need to find Fx, Fy and Fz.

Fx = 2x
Fy = -4y
Fz = -8z​

At P(x₀,y₀,z₀) on the hyperboloid, the partial derivatives will be:

Fx = 2x₀
Fy = -4y₀
Fz = -8z₀​

Putting these results in the equation of the tangent plane, we have:

2x₀(x-x₀) - 4y₀(y-y₀) – 8z₀(z-z₀) = 0
(2x₀)x – (4y₀)y – (8z₀)z = 2x₀²-4y₀²-8z₀²​

Since this plane is parallel to the given plane 4x - 2y + 4z = 5, we multiply one of the planes by any constant, say λ, to get the other one. I chose to multiply λ with the equation of the plane tangent to the hyperbpoloid:

λ [ (2x₀)x – (4y₀)y – (8z₀)z = 2x₀²-4y₀²-8z₀² ]​

Next, by comparing the respective co-efficients of x, y and z of the given and the obtained equation, we get:

2x₀ λ = 4 ------------(i)
4y₀λ = 2-------------(ii)
-8z₀λ = 4 ------------ (iii)
( 2x₀²-4y₀²-8z₀² ) λ = 5 ------------- (iv)

From (i), x₀ = 2 / λ -------------(iv)
From (ii), y₀ = 1 / 2λ ---------(v)
From (iii), z₀ = - 1 / 2λ ----------(vi)​

Putting these values in equation (iv), we get:

( 2(2/λ)²-4(1/2λ)²-8(-1/2λ)² ) λ = 5
( 8/ λ² - 1/ λ² - 2/ λ² ) λ = 5
( 5/λ² ) λ = 5
λ² - λ = 0
λ(λ-1 )= 0
λ = 0……… or ……… λ = 1​

Values of x₀, y₀ and z₀ are not possible for λ = 0, hence for λ = 1, we have from eq. (iv), (v) and (vi):

x₀ = 2
y₀ = 1/2
z₀ = -1/2​

Hence, the point on the hyperbola having the tangent plane parallel to the plane is ( 2 , 1/2 , -1/2 )

My comments:

My solution has something wrong in it. Also, this point does not satisfy the equation of the hyperboloid.

The real answers are:
( 8(sqrt2)/(sqrt5) , 2(sqrt2)/(sqrt5) , -2(sqrt2)/(sqrt5) )
and ( -8(sqrt2)/(sqrt5) , -2(sqrt2)/(sqrt5) , 2(sqrt2)/(sqrt5) )

Please help me Identify my mistake. Thanks.

 

[Stephen Tashi]

Fx = 2x
Fy = -4y
Fz = -8z​

Why did you get Fz = -8z ?

 

[HallsofIvy]

1. Homework Statement :

Find the points on the hyperboloid of two sheets with equation x²-2y²+4z²=16 at which, the tangent plane is parallel to the plane 4x-2y+4z=5.

Homework Equations

The hyperboloid with two sheets: x²-2y²+4z²=16

The given plane: 4x-2y+4z=5

Equation of tangent plane: Fx(x-x₀) + Fy(y-y₀) + Fz(z-z₀) = 0 ;
Fx, Fy and Fz are partial derivatives



3. The Attempt at a Solution :

For the hyperboloid, we can say that

F(x,y,z)=> x² - 2y² + 4z² - 16= 0​

Equation of the tangent plane is given by:

Fx(x-x₀) + Fy(y-y₀) + Fz(z-z₀) = 0​

so we need to find Fx, Fy and Fz.

Fx = 2x
Fy = -4y
Fz = -8z​

You have the sign wrong here. F_z= 8z

At P(x₀,y₀,z₀) on the hyperboloid, the partial derivatives will be:

Fx = 2x₀
Fy = -4y₀
Fz = -8z₀​

Again, sign wrong.

Putting these results in the equation of the tangent plane, we have:

2x₀(x-x₀) - 4y₀(y-y₀) – 8z₀(z-z₀) = 0
(2x₀)x – (4y₀)y – (8z₀)z = 2x₀²-4y₀²-8z₀²​

Since this plane is parallel to the given plane 4x - 2y + 4z = 5, we multiply one of the planes by any constant, say λ, to get the other one. I chose to multiply λ with the equation of the plane tangent to the hyperbpoloid:

λ [ (2x₀)x – (4y₀)y – (8z₀)z = 2x₀²-4y₀²-8z₀² ]​

Next, by comparing the respective co-efficients of x, y and z of the given and the obtained equation, we get:

2x₀ λ = 4 ------------(i)
4y₀λ = 2-------------(ii)
-8z₀λ = 4 ------------ (iii)​

Of course, that should be 8z_0\lambda= 4. You could have got these more simply by noting that the normal to the surface, \nabla x^2- 2y^2+ 16z^2= 2x\vec{i}- 4y\vec{j}+ 8z\vec{k} must be parallel to the normal to the plane, 4\vec{i}- 2\vec{j}+ 4\vec{k}.

( 2x₀²-4y₀²-8z₀² ) λ = 5 ------------- (iv)

Now, this is a major error- and the reason why your answer is not even on the hyperbola. You had the \lambda in the first three equations because the two planes only had to be parallel, not the same plane. But the point must be on[/b[] the hyperbola so you must have
2x_0^2- 4y_0^2+ 8z_0^2=5- there is no "\lambda" here.
(and note the +8, not -8). Finally, it should be equal to 16, not 5!

From (i), x₀ = 2 / λ -------------(iv)
From (ii), y₀ = 1 / 2λ ---------(v)
From (iii), z₀ = - 1 / 2λ ----------(vi)

​

z_0= 1/2\lambda

Putting these values in equation (iv), we get:

( 2(2/λ)²-4(1/2λ)²-8(-1/2λ)² ) λ = 5
( 8/ λ² - 1/ λ² - 2/ λ² ) λ = 5
( 5/λ² ) λ = 5
λ² - λ = 0​

Again, that last \lambda should not be there- you have only 5/\lambda^2= 5. And, it is equal to 16, not 5.
But even if that were correct, you have an algebra error here- (5/\lambda^2)\lambda= 5/\lambda= 5 gives 1/\lambda= 1- there is no subtraction.

[λ(λ-1 )= 0
λ = 0……… or ……… λ = 1

5/\lambda^2= 16 gives \lambda^2= 5/16 so that \lambda= \pm \sqrt{5}/4

Values of x₀, y₀ and z₀ are not possible for λ = 0, hence for λ = 1, we have from eq. (iv), (v) and (vi):

x₀ = 2
y₀ = 1/2
z₀ = -1/2​

Hence, the point on the hyperbola having the tangent plane parallel to the plane is ( 2 , 1/2 , -1/2 )

The equations should be 2x_0\lambda= 4, 4y_0\lambda= 2, and 8z_0\lambda= 4. Solve those with \lambda= \sqrt{5}/4 and \lambda= -\sqrt{5}/4.

My comments:

My solution has something wrong in it. Also, this point does not satisfy the equation of the hyperboloid.

The real answers are:
( 8(sqrt2)/(sqrt5) , 2(sqrt2)/(sqrt5) , -2(sqrt2)/(sqrt5) )
and ( -8(sqrt2)/(sqrt5) , -2(sqrt2)/(sqrt5) , 2(sqrt2)/(sqrt5) )

Please help me Identify my mistake. Thanks.

 

[gikiian]

@HallsofIvy & Stephen Tahsi:

Sir I am terribly sorry; the equation meant to be a hyperboloid in two sheets, and it it as follows:

x²-2y²-4z²=16​

So Fz=-8z.

The site which you called as the "major error" is basically a concern.

Thanks for tackling :)

P.S. I corrected the problem question above.